Robinson J C - Solution manual for An introduction to ordinary differential equations-CUP (2004)

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An Introduction to Ordinary Differential Equations
Exercises and Solutions
James C. Robinson
1
Radioactive decay and carbon dating
Exercise 1.1 Radioactive isotopes decay at random, with a fixed probability
of decay per unit time. Over a time interval ∆t, suppose that the probability
of any one isotope decaying is k∆t. If there are N isotopes, how many will
decay on average over a time interval ∆t? Deduce that
N(t + ∆t)−N(t) ≈ −Nk∆t,
and hence that dN/dt = −kN is an appropriate model for radioactive decay.
Over a time interval ∆t, Nk∆t isotopes will decay. We then have
N(t + ∆t)−N(t) = −Nk∆t.
Dividing by ∆t gives
N(t + ∆t)−N(t)
∆t
= −Nk,
and letting ∆t → 0 we obtain, using the definition of the derivative,
dN
dt
= −kN.
Exercise 1.2 Plutonium 239, virtually non-existent in nature, is one of the
radioactive materials used in the production of nuclear weapons, and is a
by-product of the generation of power in a nuclear reactor. Its half-life is
approximately 24 000 years. What is the value of k that should be used in
(1.1) for this isotope?
Since N(t) = N(s)e−k(t−s), half of the isotopes decay after a time T ,
where
N(s + T ) = 12N(s) = N(s)e
−kT ,
1
2 1 Radioactive decay and carbon dating
i.e. when 12 = e
−kT . Thus the half-life T = ln 2/k (as derived in Section
1.1). If T = 24000 then k = ln 2/T ≈ 2.888× 10−5.
Exercise 1.3 In 1947 a large collection of papyrus scrolls, including the old-
est known manuscript version of portions of the Old Testament, was found
in a cave near the Dead Sea; they have come to be known as the ‘Dead Sea
Scrolls’. The scroll containing the book of Isaiah was dated in 1994 using
the radiocarbon technique1; it was found to contain between 75% and 77%
of the initial level of carbon 14. Between which dates was the scroll written?
We have
N(1994) = pN(s) = N(s)e−k(1994−s),
where 0.75 ≤ p ≤ 0.77. Taking logarithms gives
log p = −k(1994− s),
and so
s = 1994 +
log p
k
.
With k = 1.216× 10−4 this gives (approximately)
−372 ≤ s ≤ −155,
dating the scrolls between 372 BC and 155 BC.
Exercise 1.4 A large round table hangs on the wall of the castle in Winch-
ester. Many would like to believe that this is the Round Table of King Arthur,
who (so legend would have it) was at the height of his powers in about AD
500. If the table dates from this time, what proportion of the original carbon
14 would remain? In 1976 the table was dated using the radiocarbon tech-
nique, and 91.6% of the original quantity of carbon 14 was found2. From
when does the table date?
If the table dates from 500 AD then we would expect
N(t) = e−k(t−500)N(500),
and so in 2003 we have
N(2003) = e−1503kN(500).
The proportion of 14C isotopes remaining should there be e−1503k ≈ 83%.
1 A.J. Jull et al., ‘Radiocarbon Dating of the Scrolls and Linen Fragments from the Judean
Desert’, Radiocarbon (1995) 37, 11–19.
2 M. Biddle, King Arthur’s Round Table (Boydell Press, 2001).
Radioactive decay and carbon dating 3
However, we in fact have 91.6% remaining in 1976. Therefore
N(1976) = 0.915N(s) = N(s)e−k(1993−s).
Taking logarithms gives
s = 1976 +
log 0.916
k
≈ 1255;
the table probably dates from during the reign of the English King Edward
I, who took the throne in 1270 AD (once the wood was well seasoned) and
had a passion for all things Arthurian.
Exercise 1.5 Radiocarbon dating is an extremely delicate process. Suppose
that the percentage of carbon 14 remaining is known to lie in the range 0.99p
to 1.01p. What is the range of possible dates for the sample?
Suppose that a proportion αp of the original 14C isotopes remain. Then
αpN(s) = N(t) = e−k(t−s)N(s),
and so
log α + log p = −k(t− s).
It follows that
s = t +
log p
k
+
log α
k
. (S1.1)
Denote by S the value of this expression when α = 1, i.e. S = t + (log p)/k.
For a proportion 0.99p the expression (S1.1) gives
s = S − 82.65,
while for a proportion 1.01p the expression gives
s = S + 81.83
(both correct to two decimal places). Small errors can give a difference of
over 160 years in the estimated date.
2
Integration variables
There are no exercises for this chapter.
4
3
Classification of differential equations
Exercise 3.1 Classify the following equations as ordinary or partial, give
their order, and state whether they are linear or nonlinear. In each case
identify the dependent and independent variables.
(i) Bessel’s equation (ν is a parameter)
x2y′′ + xy′ + (x2 − ν2)y = 0,
(ii) Burger’s equation (ν is a parameter)
∂u
∂t
− ν ∂
2u
∂x2
+ u
∂u
∂x
= 0,
(iii) van der Pol’s equation (m, k, a and b are parameters)
mẍ + kx = aẋ− bẋ3,
(iv) dy/dt = t− y2,
(v) the wave equation (c is a parameter)
∂2y
∂t2
= c2
∂2y
∂x2
,
(vi) Newton’s law of cooling (k is a parameter and A(t) is a specified
function)
dT
dt
= −k(T −A(t)),
(vii) the logistic population model (k is a parameter)
dp
dt
= kp(1− p),
5
6 3 Classification of differential equations
(viii) Newton’s second law for a particle of mass m moving in a potential
V (x),
mẍ = −V ′(x),
(ix) the coupled equations in (3.9)
ẋ = x(4− 2x− y)
ẏ = y(9− 3x− 3y),
and
(x)
dx
dt
= Ax,
where x is an n-component vector and A is an n× n matrix.
(i) linear 2nd order ODE for y(x);
(ii) nonlinear 2nd order PDE for u(x, t);
(iii) nonlinear 2nd order ODE for x(t);
(iv) nonlinear 1st order ODE for y(t);
(v) linear 2nd order PDE for y(x, t);
(vi) linear 1st order ODE for T (t);
(vii) nonlinear 1st order ODE for p(t);
(viii) 2nd order ODE for x(t), linear if V ′(x) = ax + b for some a, b ∈ R,
otherwise nonlinear;
(ix) nonlinear 1st order ODE for the pair (x(t), y(t)); and
(x) linear 1st order ODE for the vector x(t).
4
*Graphical representation of solutions using
MATLAB
Exercise 4.1 Plot the graphs of the following functions:
(i) y(t) = sin 5t sin 50t for 0 ≤ t ≤ 3,
(ii) x(t) = e−t(cos 2t + sin 2t) for 0 ≤ t ≤ 5,
(iii)
T (t) =
∫ t
0
e−(t−s) sin sds for 0 ≤ t ≤ 7,
(iv) x(t) = t ln t for 0 ≤ t ≤ 5,
(v) plot y against x, where
x(t) = Be−t + Ate−t and y(t) = Ae−t,
for A and B taking integer values between −3 and 3.
(i) >> t=linspace(0,3);
>> y=sin(5*t).*sin(50*t);
>> plot(t,y)
The result is shown in Figure 4.1.
(ii) >> t=linspace(0,5);
>> x=exp(-t).*(cos(2*t)+sin(2*t));
>> plot(t,x)
The result is shown in Figure 4.2.
(iii) Use the short M-file exint.m:
f=inline(’exp(-(t-s)).*sin(s)’,’t’,’s’);
for j=0:100;
t(j+1)=7*j/100;
T(j+1)=quad(f,0,t(j+1),[],[],t(j+1));
end
7
8 4 *Graphical representation of solutions using MATLAB
0 0.5 1 1.5 2 2.5 3
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Fig. 4.1. The graph of y(t) = sin 5t sin 50t against t.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Fig. 4.2. The graph of x(t) = e−t(cos 2t + sin 2t) against t.
plot(t,T)
The plot is shown in Figure 4.3.
(iv) >> t=linspace(0,5);
>> x=t.*log(t);
>> plot(t,x)
The resulting graph is shown in Figure 4.4.
(v) Use the short M-file param.m:
t=linspace(0,5);
hold on
for A=-3:3;
*Graphical representation of solutions using MATLAB 9
0 1 2 3 4 5 6 7
−200
−100
0
100
200
300
400
500
600
700
800
Fig. 4.3. The graph of the integral in Exercise 4.1(iii).
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−1
0
1
2
3
4
5
6
7
8
9
Fig. 4.4. The graph of x(t) = t ln t against t.
for B=-3:3;
x=B*exp(-t)+A.*t.*exp(-t);
y=A*exp(-t);
plot(x,y)
end
end
hold off
See Figure 4.5.
10 4 *Graphical representation of solutions using MATLAB
−3 −2 −1 0 1 2 3
−3
−2
−1
0
1
2
3
Fig. 4.5. A collection of curves defined parametrically by x(t) = Be−t + Ate−t and
y(t) = Ae−t.
Exercise 4.2 Draw contour plots of the following functions:
(i)
F (x, y) = x2 + y2 for − 2 ≤ x, y ≤ 2;
(ii)
F (x, y) = xy2 for − 1 ≤ x, y ≤ 1,
with contour lines where F = ±0.1, ±0.2, ±0.4, and ±0.8;
(iii)
E(x, y) = y2 − 2 cos x for − 4 ≤ x, y ≤ 4;
(iv)
E(x, y) = x− 13x3 + 12y2(x4 − 2x2 + 2)
for −2 ≤ x ≤ 4 and −2 ≤ y ≤ 2, showing the contour lines where
E = 0, 0.5, 0.8, 1, 2, 3, and 4;
(v)
E(x, y) = y2 + x3 − x for − 2 ≤ x, y ≤ 2.
(i) >> [x,y]=meshgrid(-2:.1:2, -2:.1:2);
>> F=x.^2+y.^2;
>> contour(x,y,F)
These contours are shown in Figure 4.6.
*Graphical representation of solutions using MATLAB 11
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Fig. 4.6. Contours of x2 + y2
(ii) >> [x, y]=meshgrid(-1:.1:1, -1:.1:1);
>> F=x.*y.^2;
>> contour(x,y,F,[.1 .2 .4 .8 -.1 -.2 -.4 -.8])
These contours are shown in Figure 4.7.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
Fig. 4.7. Contours of xy2
(iii) >> [x, y]=meshgrid(-4:.1:4, -4:.1:4);
>> E=y.^2-2*cos(x);
>> contour(x,y,E)
These contours are shown in Figure 4.8.
(iv) >> [x, y]=meshgrid(-2:.1:4, -2:.1:2);
>> E=x-(x.^3)/3+(y.^2/2).*(x.^4-2*x.^2+2);
>> contour(x,y,E,[0 .5 .8 1 2 3 4])
These contours are shown in Figure 4.9.
(v) >> [x, y]=meshgrid(-2:.1:2, -2:.1:2);
>> E=y.^2+x.^3-x;
12 4 *Graphical representation of solutions using MATLAB
−4 −3 −2 −1 0 1 2 3 4
−4
−3
−2
−1
0
1
2
3
4
Fig. 4.8. Contours of y2 − 2 cos x
−2 −1 0 1 2 3 4
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Fig. 4.9. Contours of x− 13x3 + 12y2(x4 − 2x2 + 2)
>> contour(x,y,E)
These contours are shown in Figure 4.10.
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Fig. 4.10. Contours of y2 + x3 − x
5
‘Trivial’ differential equations
Exercise 5.1 Find the general solution of the following differential equa-
tions, and in each case find also the particular solution that passes through
the origin.
(i)
dθ
dt
= sin t + cos t,
(ii)
dy
dx
=
1
x2 − 1
(use partial fractions)
(iii)
dU
dt
= 4t ln t,
(iv)
dz
dx
= xe−2x,
and
(v)
dT
dt
= e−t sin 2t.
(i) Integrating both sides of
dθ
dt
= sin t + cos t
with respect to t we get
θ(t) = − cos t + sin t + c.
13
14 5 ‘Trivial’ differential equations
For the solution to pass through the origin we need θ(0) = 0,
i.e. 0 = −1 + c or c = 1, and thus this solution is
θ(t) = 1− cos t + sin t.
(ii) We have
dy
dx
=
1
x2 − 1 =
1
2
(
1
x− 1 −
1
x + 1
)
,
and so
y(x) = 12 (log |x− 1| − log |x + 1|) + c = 12 log
|x− 1|
|x + 1| + c.
For this solution to pass through the origin we need
0 = y(0) = 12 log
| − 1|
|1| + c,
i.e. c = 0, and so we get
y(x) = 12 log
|x− 1|
|x + 1| .
(iii) Integrating
dU
dt
= 4t ln t
we have
U(t) =
∫
4t ln tdt =
1
2
t2 ln t− 1
4
t2 + c.
To ensure that U(0) = 0 we need c = 0, and so
U(t) =
1
2
t2 ln t− 1
4
t2.
(iv) Integrating the right-hand side of
dz
dx
= xe−2x
by parts we have
z(x) =
∫
xe−2x dx
= −1
2
xe−2x +
1
2
∫
e−2x dx
= −1
2
xe−2x − 1
4
e−2x + c,
‘Trivial’ differential equations 15
and for z(0) = 0 we need 0 = −(1/4) + c, i.e. c = 1/4 giving
z(x) =
1
4
(1− e−2x − 2xe−2x).
(v) We need to integrate
dT
dt
= e−t sin 2t.
The integral of e−t sin 2t will be a linear combination of e−t sin 2t and
e−t cos 2t. For
z(t) = αe−t sin 2t + βe−t cos 2t
we have
dz
dt
= α(−e−t sin 2t + 2e−t cos 2t + β(−e−t cos 2t− 2e−t sin 2t)
= (−α− 2β)e−t sin 2t + (2α− β)e−t cos 2t,
and so we need
−α− 2β = 1 and 2α− β = 0,
i.e. α = −1/5 and β = −2/5. Therefore
T (t) = −e
−t sin 2t + 2e−t cos 2t
5
+ c.
For T (0) = 0 we need 0 = −(2/5) + c, i.e. c = 2/5, and so
T (t) =
2− e−t(sin 2t + 2 cos 2t)
5
.
Exercise 5.2 Find the function f(x) defined for −π/2 < x < π/2 whose
graph passes through the point (0, 2) and has slope − tanx.
We want to find a function f that satisfies
df
dx
= − tanx with f(0) = 2.
So we integrate between the limits that correspond to x values 0 and x,
f(x)− f(0) =
∫ x
0
− tan x̃dx̃
=
∫ x
0
− sin x̃
cos x̃
dx̃
= [ln | cos x̃|]x0
= ln | cosx|,
16 5 ‘Trivial’ differential equations
and so, since cosx > 0 for −π/2 < x < π/2
f(x) = ln cosx + 2.
Exercise 5.3 Find the function g(x) defined for x > −1 that has slope
ln(1 + x) and passes through the origin.
The required function g(x) satisfies
dg
dx
= ln(1 + x) with g(0) = 0.
Integrating both sides of the differential equation between 0 and x gives
g(x) = g(0) +
∫ x
0
ln(1 + x̃) dx̃
=
[
(1 + x̃) ln(1 + x̃)− x̃
]x
x̃=0
= (1 + x) ln(1 + x)− x,
since ln 1 = 0.
Exercise 5.4 Find the solutions of the following equations satisfying the
given initial conditions:
(i)
ẋ = sec2 t with x(π/4) = 0,
(ii)
y′ = x− 13x3 with y(−1) = 1,
(iii)
dθ
dt
= 2 sin2 t with θ(π/4) = π/4,
(iv)
x
dV
dx
= 1 + x2 with V (1) = 1,
and
(v)
d
dt
[
x(t)e3t
]
= e−t with x(0) = 3,
‘Trivial’ differential equations 17
(i) Integrating ẋ = sec2 t from π/4 to t gives
x(t) = x(π/4) +
∫ t
π/4
sec2 t̃ dt̃
= 0 +
[
tan t̃
]t
t̃=π/4
= tan t− 12 .
(ii) Integrating y′ = x− 13x3 from −1 to x gives
y(x) = y(−1) +
∫ x
−1
x̃− 13 x̃3 dx̃
= 1 +
[
x̃2
2
− x̃
4
12
]x
x̃=−1
= 1 +
x2
2
− x
4
12
− 1
2
+
1
12
=
7
12
+
x2
2
− x
4
12
.
(iii) Integrating dθdt = 2 sin
2 t between π/4 and t we have
θ(t) = θ(π/4) +
∫ t
π/4
2 sin2 t̃ dt̃
= π/4 +
∫ t
π/4
1− cos 2t̃ dt̃
= π/4 +
[
t̃− 12 sin 2t̃
]t
t̃=π/4
= π/4 + t− 12 sin 2t− π/4 + 12
= 12 + t− 12 sin 2t.
(iv) Dividing xdVdx = 1 + x
2 by x and then integrating between 1 and x
we obtain
V (x) = V (1) +
∫ x
1
1
x̃
+ x̃dx̃
= 1 +
[
ln x̃ + 12 x̃
2
]x
x̃=1
= 1 + lnx + 12x
2 − ln 1− 12
= 12 + lnx +
1
2x
2.
18 5 ‘Trivial’ differential equations
(v) Integrating both sides of
d
dt
[
x(t)e3t
]
= e−t
between 0 and t gives
x(t)e3t = x(0) +
∫ t
0
e−t̃ dt̃
= 3 +
[
−e−t̃
]t
t̃=0
= 3− e−t + 1
= 4− e−t,
and so
x(t) = 4e−3t − e−4t.
Exercise 5.5 The Navier-Stokes equations that govern fluid flow were given
as an example in Chapter 3 (see equations (3.1) and (3.2)). It is not possible
to find explicit solutions of these equations in general. However, in certain
cases the equations reduce to something much simpler.
Suppose that a fluid is flowing down a pipe that has a circular cross-section
of radius a. Assuming that the velocity V of the fluid depends only on its
distance from the centre of the pipe, the equation satisfied by V is
1
r
d
dr
(
r
dV
dr
)
= −P,
where P is a positive constant.
Multiply by r and integrate once to show that
dV
dr
= −Pr
2
+
c
r
where c is an arbitrary constant. Integrate again to find an expression for
the velocity, and then use the facts that (i) the velocity should be finite at
all points in the pipe and (ii) that fluids ‘stick’ to boundaries (which means
that V (a) = 0) to show that
V (r) =
P
4
(a2 − r2),
see Figure 5.1. (This is known as Poiseuille flow.)
‘Trivial’ differential equations 19
a
 V(r)=P(a2−r2)/4
Fig. 5.1. The quadratic velocity profile in a circular pipe.
Multiplying the equation by r we obtain
d
dr
(
r
dV
dr
)
= −Pr,
and integrating both sides gives
r
dV
dr
= −Pr
2
2
+ c,
which implies that
dV
dr
= −Pr
2
+
c
r
.
Integrating this equation gives
V (r) = −Pr
2
4
+ c ln r + d.
Since ln r → −∞ as r → 0, for V (r) to be finite when r = 0 we have to take
c = 0. This then leaves
V (r) = −Pr
2
4
+ d,
and to ensure that V (a) = 0 we take d = Pa2/4, so that
V (r) =
P
4
(a2 − r2)
as claimed.
Exercise 5.6 An apple of mass m falls from a height h above the ground.
Neglecting air resistance its velocity satisfies
m
dv
dt
= −mg v(0) = 0,
20 5 ‘Trivial’ differential equations
where v = ẏ and y is the height above ground level. Show that the apple hits
the ground when
t =
√
2h
g
.
The velocity at time t is given by
v(t) = v(0)− gt = −gt,
and its height y above ground level satisfies
ẏ = v(t) = −gt,
and hence
y(t) = y(0)− 12gt2 = h− 12gt2.
It follows that y(t) = 0 when t =
√
2h/g as claimed.
Exercise 5.7 An artillery shell is fired from a gun, leaving the muzzle with
velocity V . If the gun is at an angle θ to the horizontal then the initial
horizontal velocity is V cos θ, and the initial vertical velocity is V sin θ (see
Figure 5.2). The horizontal velocity remains constant, but the vertical ve-
locity is affected by gravity, and obeys the equation v̇ = −g. How far does
the shell travel before it hits the ground? (Give your answer in terms of V
and θ.)
V
θ
Fig. 5.2. Firing a shell at muzzle velocityV at an angle θ to the horizontal. The
shell follows a parabolic path.
The vertical velocity v satisfies
v̇ = −g with v(0) = V sin θ,
and so integrating both sides of the differential equation between times 0
and t we obtain
v(t) = v(0)− gt = V sin θ − gt.
The height y(t) of the shell at time t satisfies
ẏ = v = V sin θ − gt with y(0) = 0,
‘Trivial’ differential equations 21
and so integrating both sides of this between zero and t we have
y(t) = V t sin θ − 12gt2.
The shell strikes the ground at time t∗, where
V t∗ sin θ − 12gt2∗ = 0,
i.e. when t∗ = (2V sin θ)/g. Since the horizontal velocity is constant and
equal to V cos θ, the shell will have travelled a distance
V t∗ cos θ =
2V 2 sin θ cos θ
g
=
V 2 sin 2θ
g
.
Exercise 5.8 In Dallas on 22 November 1963, President Kennedy was as-
sassinated; by Lee Harvey Oswald if you do not believe any of the conspiracy
theories. Oswald fired a Mannlicher-Carcano rifle from approximately 90 m
away. The sight on Oswald’s rifle was less than ideal: if the bullet travelled
in a straight line after leaving the rifle (at a velocity of roughly 700 m/s)
then the sight aimed about 10cm too high at a target 90 m away. How much
would the drop in the trajectory due to gravity compensate for this? (The
initial vertical velocity v is zero, and satisfies the equation v̇ = −g, while the
horizontal velocity is constant if we neglect air resistance.)
There is nothing to slow down the horizontal velocity of the bullet if we
neglect air resistance: so it takes the bullet 9/70 seconds to travel 90 m. In
this time it will have dropped vertically, its height h satisfying
d2h
dt2
= −g.
The solution of this, integrating twice, is
h(t) = h(0)− 12gt2,
and so with t = 9/70 and h(0) = 0 this gives a drop of 0.081 m or 8.1 cm,
compensating quite well for the dodgy sight.
Exercise 5.9 This exercise fills in the gaps in the proof of the Fundamental
Theorem of Calculus. Suppose that f is continuous at x, i.e. given any
² > 0, there exists a δ = δ(²) such that
|x̃− x| ≤ δ ⇒ |f(x̃)− f(x)| ≤ ².
By writing
f(x) =
1
δx
∫ x+δx
x
f(x) dx̃
22 5 ‘Trivial’ differential equations
show that for all δx with |δx| ≤ δ(²)
∣∣∣∣f(x)−
1
δx
∫ x+δx
x
f(x̃) dx̃
∣∣∣∣ ≤ ²,
and hence that
lim
δx→0
1
δx
∫ x+δx
x
f(x̃) dx̃ = f(x).
You will need to use the fact that
∣∣∣∣
∫ b
a
g(x) dx
∣∣∣∣ ≤
∫ b
a
|g(x)|dx ≤ (b− a) max
x∈[a,b]
|g(x)|.
We have
∣∣∣∣f(x)−
1
δx
∫ x+δx
x
f(x̃) dx̃
∣∣∣∣ =
∣∣∣∣
1
δx
∫ x+δx
x
f(x) dx̃− 1
δx
∫ x+δx
x
f(x̃) dx̃
∣∣∣∣
=
1
δx
∣∣∣∣
∫ x+δx
x
f(x)− f(x̃) dx̃
∣∣∣∣
≤ 1
δx
∫ x+δx
x
|f(x)− f(x̃)| dx̃.
Then, given any ² > 0, there exists a δ(²) > 0 such that if δx < δ(²) then
for every x̃ ∈ [x, x + δx] we have |f(x)− f(x̃)| ≤ ², and so
∣∣∣∣f(x)−
1
δx
∫ x+δx
x
f(x̃) dx̃
∣∣∣∣ ≤
1
δx
∫ x+δx
x
²dx̃
=
1
δx
[² δx]
= ².
Therefore, using the definition of a limit,
lim
δx→0
1
δx
∫ x+δx
x
f(x̃) dx̃ = f(x),
as claimed.
6
Existence and uniqueness of solutions
Exercise 6.1 Which of the following differential equations have unique so-
lutions (at least on some small time interval) for any non-negative initial
condition (x(0) ≥ 0)?
(i) ẋ = x(1− x2)
(ii) ẋ = x3
(iii) ẋ = x1/3
(iv) ẋ = x1/2(1 + x)2
(v) ẋ = (1 + x)3/2.
In each of these questions we will denote by f(x) the right-hand side of
the differential equation. We need to check whether or not f and f ′ are
continuous for x ≥ 0.
(i) Here f(x) = x(1 − x2) and f ′(x) = 1 − 3x2 are both continuous, so
solutions are unique. [In fact for x(0) ≥ 0 solutions exist for all t ∈ R,
while for x(0) < 0 they ‘blow up’ to x = −∞ in a finite time.]
(ii) For this example f(x) = x3 and f ′(x) = 3x2 so solutions are unique.
[Solutions blow up in a finite time unless x(0) = 0.]
(iii) We have f(x) = x1/3 (which is continuous), but f ′(x) = x−2/3/3, so
f ′(x) →∞ as x → 0, and the solution of
ẋ = x1/3 with x(0) = 0
is not unique: for any choice of c ≥ 0, the function
x(t) =



0 t < c(
2(t−c)
3
)3/2
t ≥ c
solves this equation. [Note that unlike the example ẋ = x1/2 this
equation also makes sense for x < 0.]
23
24 6 Existence and uniqueness of solutions
(iv) We have f(x) = x1/2(1 + x2) which is continuous, and
f ′(x) = 12x
−1/2(1 + x2) + 2x3/2.
Near zero f ′(x) → ∞, and so the solutions with x(0) = 0 are not
unique.
(v) The function f(x) = (1+x)3/2 is continuous, and f ′(x) = 32(1+x)
1/2
is also continuous, so solutions are unique. [Solutions blow up in a
finite time.]
Exercise 6.2 The Mean Value Theorem says that if f is differentiable on
an interval [a, b] then f(a)−f(b) = (b−a)f ′(c) for some c ∈ (a, b). Suppose
that f(x) is differentiable with |f ′(x)| ≤ L for a ≤ x ≤ b. Use the Mean
Value Theorem to show that for a ≤ x, y ≤ b we have
|f(x)− f(y)| ≤ L|x− y|.
The result is clearly true if x = y. Using the mean value theorem for
x > y we have
f(x)− f(y) = f ′(c)(x− y)
for some c ∈ (x, y). It follows that
|f(x)− f(y)| ≤ |f ′(c)||x− y|,
and since |f ′(c)| ≤ L we have
|f(x)− f(y)| ≤ L|x− y|. (S6.1)
If y > x then f(y)− f(x) = f ′(c)(y− x) and on taking the modulus of both
sides we once again arrive at (S6.1).
Exercise 6.3 This Exercise gives a simple proof of the uniqueness of solu-
tions of
ẋ = f(x, t) x(t0) = x0, (S6.2)
under the assumption that
|f(x, t)− f(y, t)| ≤ L|x− y|. (S6.3)
Suppose that x(t) and y(t) are two solutions of (S6.2). Write down the
differential equation satisfied by z(t) = x(t)− y(t), and hence show that
d
dt
|z|2 = 2z[f(x(t), t)− f(y(t), t)].
Existence and uniqueness of solutions 25
Now use (S6.3) to show that
d
dt
|z|2 ≤ 2L|z|2.
If dZ/dt ≤ cZ it follows that Z(t) ≤ Z(t0)ec(t−t0) (see Exercise 9.7): use
this to deduce that the solution of (S6.2) is unique. Hint: any two solutions
of (S6.2) agree when t = t0.
We have
dx
dt
= f(x, t)
and
dy
dt
= f(y, t).
It follows that
dz
dt
=
d
dt
(x− y) = dx
dt
− dy
dt
= f(x, t)− f(y, t).
Now,
d
dt
|z|2 = dz
dt
2
= 2z
dz
dt
= 2z[f(x, t)− f(y, t)].
Using the Lipschitz property (S6.3) we have
f(x, t)− f(y, t) ≤ |f(x, t)− f(y, t)| ≤ L|x− y| = L|z|,
and so
d
dt
|z|2 ≤ 2Lz|z| ≤ 2L|z|2.
It follows (using dZ/dt ≤ cZ ⇒ Z(t) ≤ Z(t0)ec(t−t0)) that
|z(t)|2 ≤ |z(t0)|2e2L(t−t0). (S6.4)
Since x(t0) = y(t0) we have z(t0) = 0, and so (S6.4) becomes |z(t)|2 = 0. It
follows that z(t) = 0, and so x(t) = y(t), which shows that the two solutions
must be identical.
Exercise 6.4 (T) The proof of existence of solutions is much more involved
than the proof of their uniqueness. We will consider here the slightly simpler
case
ẋ = f(x) with x(0) = x0, (S6.5)
assuming that
|f(x)− f(y)| ≤ L|x− y|. (S6.6)
26 6 Existence and uniqueness of solutions
The first step is to convert the differential equation into an integral equation
that is easier to deal with: we integrate both sides of (S6.5) between times 0
and t to give
x(t) = x0 +
∫ t
0
f(x(t̃)) dt̃. (S6.7)
This integral equation is equivalent to the original differential equation: any
solution of (S6.7) will solve (S6.5), and vice versa.
The idea behind the method is to use the right-hand side of (S6.7) as a
means of refining any ‘guess’ of the solution xn(t) by replacing it with
xn+1(t) = x0 +
∫ t
0
f(xn(t̃)) dt̃. (S6.8)
We start with x0(t) = x0 for all t, set
x1(t) = x0 +
∫ t
0
f(x0) dt̃,
and continue in this way using (E6.6). The hope is that xn(t) will converge
to the solution of the differential equation as n →∞.
(i) Use (S6.6) to show that
|xn+1(t)− xn(t)| ≤ L
∫ t
0
|xn(t̃)− xn−1(t̃)| dt̃,
and deduce that
max
t∈[0,1/2L]
|xn+1(t)− xn(t)| ≤ 12 maxt∈[0,1/2L] |xn(t)− xn−1(t)|. (S6.9)
(ii) Using (S6.9) show that
max
t∈[0,1/2L]
|xn+1(t)− xn(t)| ≤ 12n−1 maxt∈[0,1/2L] |x1(t)− x0(t)|. (S6.10)
(iii) By writing
xn(t) = [xn(t)−xn−1(t)]+[xn−1(t)−xn−2(t)]+· · ·+[x1(t)−x0(t)]+x0(t)
deduce that
max
t∈[0,1/2L]
|xn(t)− xm(t)| ≤ 12N−2 maxt∈[0,1/2L] |x1(t)− x0(t)|
for all n,m ≥ N .
Existence and uniqueness of solutions 27
It follows that xn(t) converges to some function x∞(t) as n → ∞, and
therefore taking limits in both sides of (E6.6) impliesthat
x∞(t) = x0 +
∫ t
0
f(x∞(t̃)) dt̃.
Thus x∞(t) satisfies (S6.7), and so is a solution of the differential equation.
The previous Exercise shows that this solution is unique.
(i) We have
xn+1(t)− xn(t) = x0 +
∫ t
0
f(xn(t̃)) dt̃− x0 −
∫ t
0
f(xn−1(t̃)) dt̃
=
∫ t
0
f(xn(t̃))− f(xn−1(t̃)) dt̃,
and so
|xn+1(t)− xn(t)| =
∣∣∣∣
∫ t
0
f(xn(t̃))− f(xn−1(t̃)) dt̃
∣∣∣∣
≤
∫ t
0
|f(xn(t̃))− f(xn−1(t̃))| dt̃
≤
∫ t
0
L|xn(t̃)− xn−1(t̃)| dt̃,
= L
∫ t
0
|xn(t̃)− xn−1(t̃)| dt̃,
using (S6.6).
Since, therefore,
|xn+1(t)− xn(t)| ≤ Lt max
t̃∈[0,t]
|xn(t̃)− xn−1(t̃)|
≤ L 1
2L
max
t̃∈[0,1/2L]
|xn(t̃)− xn−1(t̃)|
for any t ∈ [0, 1/2L], it follows that
max
t∈[0,1/2L]
|xn+1(t)− xn(t)| ≤ 12 maxt∈[0,1/2L] |xn(t)− xn−1(t)|, (S6.11)
as claimed.
(ii) We will write
Dn = max
t∈[0,1/2L]
|xn(t)− xn−1(t)|;
28 6 Existence and uniqueness of solutions
then (S6.11) reads Dn+1 ≤ 12Dn. It follows easily that
Dn+1 ≤ 2−(n−1)D1
which is (S6.10).
(iii) Taking (wlog) n > m we have
xn(t)− xm(t) = xn(t)− xn−1(t) + xn−1(t)− xn−2(t)
+ . . . + xm+1(t)− xm(t),
it follows that
max
t∈[0,1/2L]
|xn(t)− xm(t)| ≤ Dn + Dn−1 + . . . + Dm+1
≤
[
1
2n−2
+
1
2n−3
+ . . . +
1
2m−1
]
D1
≤ D1
2m−2
.
7
Scalar autonomous ODEs
Exercise 7.1 For each of the following differential equations draw the phase
diagram, labelling the stationary points as stable or unstable.
(i) ẋ = −x + 1
(ii) ẋ = x(2− x)
(iii) ẋ = (1 + x)(2− x) sin x
(iv) ẋ = −x(1− x)(2− x)
(v) ẋ = x2 − x4
The Figures all show the phase diagram and the graph of the function
f(x) on the right-hand side of the equation.
(i)
ẋ = −x + 1.
There is a single stationary point at x = 1, as shown in Figure 7.1.
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
S
Fig. 7.1. The phase diagram for ẋ = −x + 1.
29
30 7 Scalar autonomous ODEs
(ii)
ẋ = x(2− x).
There are two stationary points, x = 0 and x = 2, as shown in Figure
7.2.
−1 −0.5 0 0.5 1 1.5 2 2.5 3
−1.5
−1
−0.5
0
0.5
1
U S
Fig. 7.2. The phase diagram for ẋ = x(2− x).
(iii)
ẋ = (1 + x)(2− x) sinx.
There are an infinite number of stationary points, x = −1, x = 2,
and x = nπ for any integer n, as shown in Figure 7.3.
−8 −6 −4 −2 0 2 4 6 8
−30
−25
−20
−15
−10
−5
0
5
10
15
20
S U S U S U S
Fig. 7.3. The phase diagram for ẋ = (1 + x)(2− x) sin x for −7 ≤ x ≤ 7. There are
other stationary points at x = nπ for any integer n.
(iv)
ẋ = −x(1− x)(2− x).
There are three stationary points, x = 0, x = 1, and x = 2, as shown
in Figure 7.4.
(v)
ẋ = x2 − x4.
Scalar autonomous ODEs 31
−0.5 0 0.5 1 1.5 2 2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
S U S
Fig. 7.4. The phase diagram for ẋ = −x(1− x)(2− x).
There three are stationary points, at x = 0, x = 1, and x = −1, as
shown in Figure 7.5.
−1.5 −1 −0.5 0 0.5 1 1.5
−1
−0.5
0
0.5
U U S
Fig. 7.5. The phase diagram for ẋ = x2 − x4.
Exercise 7.2 For the equations in Exercise 7.1 determine the stability of
the stationary points analytically, by considering the sign of the derivative
of the right-hand side.
(i) When f(x) = −x + 1 we have f ′(x) = −1, and the stationary point
at x = 1 is stable.
(ii) When f(x) = x(2− x) we have f ′(x) = 2− 2x: we have f ′(0) = 2, so
that the stationary point x = 0 is unstable; and we have f ′(2) = −2,
and this stationary point is stable.
32 7 Scalar autonomous ODEs
(iii) For f(x) = (1 + x)(2− x) sin x we have
f ′(x) = (1− 2x) sin x + (1 + x)(2− x) cos x.
So at x = −1 we have
f ′(−1) = 3 sin(−1) ≈ −2.52 < 0
and this point is stable; at x = 2 we have
f ′(2) = −3 sin 2 ≈ −2.73 < 0
and this point is also stable.
At x = nπ we have
f ′(nπ) = (1 + nπ)(2− nπ)(−1)n;
taking n = 0 gives f ′(0) = 2, and this point is unstable. For integer
n 6= 0,
(1 + nπ)(2− nπ) ≤ 0,
and so f ′(nπ) > 0 (these points are unstable) if n is odd and f ′(nπ) <
0 if n is even (and these points are stable).
(iv) We have f(x) = −x(1− x)(2− x), and so
f ′(x) = −3x2 + 6x− 2.
Therefore
f ′(0) = −2, f ′(1) = 1, and f ′(2) = −2,
and so x = 0 and x = 2 are stable, while x = 1 is unstable.
(v) Now we have f(x) = x2 − x4, and so f ′(x) = 2x− 4x3. This gives
f ′(−1) = 2, f ′(0) = 0, and f ′(1) = −2.
We can only tell using this method that x = −1 is unstable and
that x = 1 is stable. To find the stability of x = 0 we need to work
from the phase diagram (it is unstable, or, if you prefer, ‘semi-stable’,
i.e. stable on one side and unstable on the other).
Exercise 7.3 For all positive values of c find all the stationary points of
dx
dt
= sin x + c,
and determine analytically which are stable and unstable. Draw the portion
of the phase diagram between −π and π. There are three different cases,
Scalar autonomous ODEs 33
0 ≤ c < 1, c = 1, and c > 1. You will need to be more careful with the case
c = 1.
Stationary points occur whenever
f(x) ≡ sinx + c
is zero, so whenever sinx = −c. If 0 ≤ c < 1 then there are two solutions
in (−π, π], x1 between −π and −π/2, and x2 between −π/2 and 0. Since
f ′(x) = cosx we have
f ′(x1) < 0 and f ′(x2) > 0,
so that x1 is stable and x2 is unstable, see Figure 7.6.
−3 −2 −1 0 1 2 3
−0.5
0
0.5
1
1.5
US
Fig. 7.6. The two stationary points in the interval [−π, π] for the equation ẋ =
sin x + c when 0 ≤ c < 1. This figure has c = 1/2.
When c = 1 the equation sinx = −1 has only one solution between −π
and π, which is x = −π/2. Since f ′(−π/2) = cos(−π/2) = 0 we need to look
more closely. We have f ′′(x) = − sinx, so at −π/2 the second derivative is
positive. So f(x) is a minimum when x = −π/2. It follows that f(x) itself
is positive on both sides on x = −π/2, so the stationary point is unstable
(“stable from the left” but “unstable to the right”), as in Figure 7.7.
Finally when c > 1 there are no stationary points, and f(x) > 0 for all x.
So all trajectories move to the right, as in Figure 7.8.
Exercise 7.4 A simple model of the spread of an infection in a population
is
Ḣ = −kIH
İ = kIH,
34 7 Scalar autonomous ODEs
−3 −2 −1 0 1 2 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
U
Fig. 7.7. The phase portrait near the single stationary point x = −π/2 for the
equation ẋ = sin x + 1.
−3 −2 −1 0 1 2 3
0
0.5
1
1.5
2
2.5
Fig. 7.8. The phase portrait for the equation ẋ = sin x + c when c > 1 (this figure
has c = 1.5). There are no stationary points.
where H(t) is the number of healthy people, I(t) the number of infected people
and k the rate of infection. Since (d/dt)(H + I) = 0, it follows that the size
of the population is constant, H + I = N , say. Substitute I = N − H in
order to obtain a single equation for H(t),
dH
dt
= −kH(N −H).
Determine the stability of the stationary points for this equation, and draw its
phase diagram. Deduce that eventually all the population becomes infected.
We have Ḣ = −kIH and I = N −H. It follows that
dH
dt
= −kH(N −H).
Scalar autonomous ODEs 35
The stationary points are H = 0 (everybody infected) and H = N (nobody
infected). If f(H) = −kH(N −H) then f ′(H) = −kN +2kH; since f ′(0) =
−kN and f ′(N) = kN it follows that H = 0 is stable and H = N is
unstable. The phase diagram is shown in Figure 7.9; the whole population
is eventually infected.
Stable Unstable
0 N
Fig. 7.9. The phase diagram for Ḣ = −KH(N − H): eventually there are no
healthy members of the population left.
Exercise 7.5 Consider the equation
dx
dt
= f(x) ≡ x2 − k.
Draw the phase diagram for the three cases k < 0, k = 0 and k > 0, labelling
the stationary points as stable or unstable in each case. Find the stability
of the stationary points using an analytic method when k > 0. Show that
f ′(0) = 0 when k = 0. Why is this significant?
Draw the bifurcation diagram, with k on the horizontal axis and the fixed
points plotted against k, indicating stable fixed points by a solid line and
unstable fixed points by a dashed line. (This is known as a saddle node
bifurcation.)
For ẋ = x2−k when k < 0 there are no stationary points and the particle
always moves to the right, as in Figure 7.10.
When k = 0 the equation is ẋ =x2: there is one stationary point at x = 0;
we have finite-time blowup for x(0) > 0, see Figure 7.11.
However, when k > 0 there are two stationary points at ±
√
k, one stable
and one unstable, as shown in Figure 7.12.
We can check the stability and instability of the stationary points for
k > 0 analytically, since f ′(x) = 2x: f ′(−
√
k) = −2
√
k and so this point is
stable, while f ′(
√
k) = 2
√
k and this point is unstable.
36 7 Scalar autonomous ODEs
Fig. 7.10. The phase diagram for ẋ = x2 − k when k < 0. There are no stationary
points.
U
0
Fig. 7.11. The phase diagram for ẋ = x2. There is one stationary point at zero.
When k = 0 we have f ′(0) = 0. This is a necessary condition for a
bifurcation, which is precisely what we have here (the so-called saddle node
bifurcation). Plotting the stationary points on a graph of (x vs k), with
the stable points shown as a solid line and the unstable points as a dashed
line we have the bifurcation diagram shown in Figure 7.13 (cf. the pitchfork
bifurcation in section 7.6). In this ‘saddle-node bifurcation’ two stationary
points (a stable point and an unstable point) appear ‘from nowhere’.
Exercise 7.6 Draw the phase diagram for the equation
ẋ = g(x) = kx− x2
for k < 0, k = 0, and k > 0. Check the stability of the stationary points by
considering g′(x), and show that the two stationary points exchange stability
as k passes through zero. Draw the bifurcation diagram for this transcritical
bifurcation.
The stationary points occur where kx − x2 = 0, i.e. at x = 0 and x = k.
Figure 7.14 shows the phase diagrams (along with the graph of f) for k < 0,
k = 0, and k > 0.
Scalar autonomous ODEs 37
S U
−√k √k
Fig. 7.12. The phase diagram for ẋ = x2 − k when k is positive. There are two
stationary points, a stable point at x = −
√
k and an unstable point at x =
√
k.
−0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
parameter k
st
at
io
na
ry
 p
oi
nt
s
Fig. 7.13. The bifurcation for the ‘saddle-node bifurcation’ that occurs in the equa-
tion ẋ = x2 − k.
With g(x) = kx − x2 we have g′(x) = k − 2x, and so g′(0) = k and
g′(k) = −k. It follows that for k < 0, the point x = 0 is stable and
x = k is unstable, while for k > 0 this stability is reversed, with x = 0
becoming unstable and x = k becoming stable. When k = 0 we have
only one stationary point, at x = 0, and there g′(0) = 0; the stability is
indeterminate and their is the possibility of a bifurcation. The bifurcation
diagram is shown in Figure 7.15.
Exercise 7.7 One equation can exhibit a number of bifurcations. Find,
depending on the value of k, all the stationary points of the equation
ẋ = h(x) = −(1 + x)(x2 − k)
38 7 Scalar autonomous ODEs
k 0
0
0 k
Fig. 7.14. The phase diagram for ẋ = kx− x2 for (left to right) k < 0, k = 0, and
k > 0.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
k
st
at
io
na
ry
 p
oi
nt
s
Fig. 7.15. A transcritical bifurcation: stability is exchanged between the origin and
x = k as k passes through zero. A solid line indicates a stable stationary point,
and a dashed line an unstable stationary point
and by considering h′(x) determine their stability. At which points, and for
which values of k, are there possible bifurcations?
Draw representative phase portraits for the five distinct parameter ranges
k < 0, k = 0, 0 < k < 1, k = 1, and k > 1, and then draw the bifurcation
diagram. Identify the type of the two bifurcations.
The stationary points occur when
h(x) = −(1 + x)(x2 − k) = 0,
so at x = −1 and x = ±
√
k. For k < 0 there is only one stationary point
(x = −1), for k = 0 there are two (x = −1 and x = 0), and for k > 0
there are three, x = −1 and x = ±
√
k, apart from k = 1 when the points at
x = −1 and x = −
√
k coincide. Since
h′(x) = −3x2 − 2x + k
Scalar autonomous ODEs 39
we have
h′(−1) = k − 1 and h′(±
√
k) = −2(k ±
√
k).
So x = −1 is stable for k < 1 and unstable for k > 1. When k = 1 we
have h′(−1) = 0, and there may be a bifurcation near this point.
When k = 0 we have h′(0) = 0, meaning that there may be a bifurcation
near zero. For k > 0 we always have h′(
√
k) < 0 so that this point is
always stable. For k < 1 we have h′(−
√
k) > 0 implying that x = −
√
k is
unstable, but for k > 1 we have h′(−
√
k) < 0 giving stability; for k = 1 the
derivative takes the indeterminate value h′(−
√
k) = 0, giving the chance of a
bifurcation (this again indicates the possibility of a bifurcation near x = −1
when k = 1, as above).
The phase portraits for k < 0, k = 0, 0 < k < 1, k = 1, and k > 1,
are shown in Figure 7.17. The bifurcation diagram is shown in Figure 7.16.
The bifurcation at k = 0 is a saddle-node bifurcation (two new stationary
points appear), and that at k = 1 is a transcritical bifurcation (a transfer of
stability).
−0.5 0 0.5 1 1.5 2
−1.5
−1
−0.5
0
0.5
1
1.5
k
st
at
io
na
ry
 p
oi
nt
s
Fig. 7.16. The bifurcation diagram for the equation ẋ = −(1 + x)(x2 − k)
[solid/dashed lines indicate stable/unstable stationary points]. There is a saddle-
node bifurcation near x = 0 as k passes through zero, and a transcritical bifurcation
near x = −1 and k passes through one.
In the remaining exercises assume that f is a C1 function, i.e. that both
f and df/dx are continuous functions. Note that such an f is smooth
enough to guarantee that the equation ẋ = f(x) with x(t0) = x0 has a
unique solution. You may also assume that the solutions are defined for all
t ≥ 0.
40 7 Scalar autonomous ODEs
−1
−1 0
−1 −√k √k
−1 −1
−1−√k √k
Fig. 7.17. Phase portraits for the equation ẋ = −(1 + x)(x2 − k) for, from top to
bottom: k < 0, k = 0, 0 < k < 1, k = 1, and k > 1.
Exercise 7.8 Let x(t) be one solution of the differential equation
ẋ = f(x).
Show that
(i) if f(x(t∗)) = 0 for some t∗ then x(t) = x(t∗) for all t ∈ R (the
solution is constant, and x(t∗) is a stationary point); and hence
Scalar autonomous ODEs 41
(ii) if f(x(t∗)) > 0 for some t∗ then f(x(t)) > 0 for all t ∈ R (the solution
cannot ‘reverse direction’). Hint: Use the Intermediate Value The-
orem, which states that if g is a continuous function with g(a) < 0
and g(b) > 0 then there is a point c between a and b with g(c) = 0.
Of course, a similar result to (ii) holds if f(x(t∗)) < 0 for some t∗.
(i) If f(x(t∗)) = 0 then write x∗ = x(t∗). Now, since f(x∗) = 0, it
follows that x(t) = x∗ for all t ∈ R is a solution of the equation
ẋ = f(x). Since solutions are unique, there is only one solution with
x(t∗) = x∗, and so this must be x(t) = x∗ for all t. So we must have
x(t) = x(t∗) for all t, as claimed. This shows that if a solution is ever
at a stationary point, then it is always there.
(ii) First note that since f and x are continuous, f(x(t)) is a continuous
function of t. We know that f(x(t∗)) > 0; now suppose that f(x(s)) <
0 for some s. It follows from the intermediate value theorem that
there must be some τ between t∗ and s with f(x(τ)) = 0. But, then,
by part (i), the solution must in fact be constant with x(t) = x(τ)
for all t, a contradiction. So f(x(t)) > 0 for all t.
Exercise 7.9 Show that for autonomous scalar equations, if x∗ is attracting
then it must also be stable. Hint: use (ii) above.
If x∗ is attracting then there exists some ² > 0 such that
|x0 − x∗| < ² ⇒ |x(t)− x∗| → 0 as t →∞.
Suppose that x0 < x∗ (a similar argument works if x0 > x∗). Then since
the solution has to move towards x∗, at some time we must have x(t) > x0,
and at some time t∗ > 0 we must have f(x(t∗)) > 0. But then, by part (ii)
of the previous question, the solution x(t) must have f(x(t)) > 0 for all t,
so be increasing for all t. It follows that
|x0 − x∗| < ² ⇒ |x(t)− x∗| < ² for all t ≥ 0,
and so x∗ is stable.
Exercise 7.10 Suppose that x(t) is a solution of ẋ = f(x) that is moving
to the right. Show that either x(t) → +∞, or x(t) → x∗, where x∗ is a
stationary point. (Hint: If x(t) does not tend to infinity then it is increasing
and bounded above, andso tends to a limit x∗. Show that in this case we
must have f(x∗) = 0.) A similar result holds if x(t) is moving to the left,
with +∞ replaced by −∞.
42 7 Scalar autonomous ODEs
Following the hint, either x(t) → +∞, or x(t) is an increasing function
that is bounded above. In the latter case, it follows that x(t) → x∗ for some
x∗. Suppose that f(x∗) 6= 0. If f(x∗) < 0 then f(x) < 0 for x sufficiently
close to x∗, which contradicts the fact that x(t) must always move right.
If f(x∗) > 0 then, since f is continuous, for some ² > 0 we know that
f(x) > 12f(x
∗) for all x ∈ (x∗− ², x∗ + ²). Since x(t) → x∗, for some T large
enough x(T ) > x∗ − ². But then
dx
dt
(t) = f(x(t)) > 12f(x
∗) > 0,
and integrating this from T to t gives x(t) > x(T ) + 12(t − T )f(x∗). This
shows that for t large enough x(t) > x∗, contradicting the fact that x(t) ↑ x∗.
Exercise 7.11 Suppose that ẋ = f(x) has a stable stationary point at x0,
with f(x0) < 0. Let g be another C1 function. Use the following scalar
version of the Implicit Function Theorem to show that for ² sufficiently small
the equation
ẋ = f(x) + ²g(x)
has a unique stationary point near x0 which is still stable.
Theorem. Suppose that h(x, ²), ∂h/∂x, ∂h/∂² are all continuous functions
of both x and ². Suppose also that h(x0, 0) = 0 and ∂h/∂x(x0, 0) 6= 0. Then
there is an open interval I that contains x0 such that for each ² sufficiently
small there is a unique solution y(²) ∈ I of
h(y(²), ²) = 0,
and y(²) depends continuously on ².
To use the theorem set
h(x; , ²) = f(x) + ²g(x).
Then h, ∂h/∂x = f ′(x) + ²g′(x), and ∂h/∂² = g(x) are all continuous
functions of ² and x.
Clearly h(x∗, 0) = f(x∗) = 0 at the stationary point x∗, while from our
assumption that f ′(x∗) < 0 we have ∂h/∂x(x∗, 0) = f ′(x∗) < 0.
Then in some open interval containing x∗ (i.e. (x∗ − δ, x∗ + δ) for some
small δ) there is a unique solution y of the equation
f(y) + ²g(y) = 0,
i.e. a unique stationary point of the equation
ẋ = f(x) + ²g(x).
Scalar autonomous ODEs 43
Since f ′ and g′ are continuous functions of x, and y depends continuously
on ², it follows for ² sufficiently small that
f ′(y) + ²g′(y) < 0,
and so y is still stable.
8
Separable equations
Exercise 8.1 Solve the following equations:
(i) ẋ = t3(1− x) with x(0) = 3;
(ii) y′ = (1 + y2) tan x with y(0) = 1;
(iii) ẋ = t2x (general solution);
(iv) ẋ = −x2 (general solution);
(v) for dy/dt = e−t2y2 give the solution in terms of an integral and
describe the behaviour of the solution as t → +∞, depending on the
initial condition. You may assume that
∫∞
0 e
−s2 ds =
√
π/2.
(i) Separate variables in the equation dx/dt = t3(1− x) to give
dx
1− x = t
3 dt
and integrate between the limits associated with t values 0 and t,
∫ x(t)
3
dx
1− x =
∫ t
0
t̃3 dt̃
to give
[
− ln |1− x|
]x(t)
x=3
=
[
t̃4
4
]t
t̃=0
.
Therefore
− ln |1− x(t)|+ ln 2 = t
4
4
.
Exponentiating both sides gives
2
|x(t)− 1| = e
t4/4
44
Separable equations 45
and rearranging we have
|x(t)− 1| = 2e−t4/4.
Since x(0) = 3 for small t we have y(t)− 1 > 0, so
y(t) = 1 + 2e−t
4/4.
Noting that this gives y(t)−1 > 0 for all t this is indeed the required
solution.
(ii) Separating the variables in the equation dy/dx = (1 + y2) tan x we
have
dy
1 + y2
= tanxdx,
and so integrating both sides between the limits associated with x
values 0 and x we get
∫ y(x)
1
dy
1 + y2
=
∫ x
0
tan x̃dx̃
which gives
[
tan−1 y
]y(x)
y=1
=
[
− ln cos x̃
]x
x̃=0
.
Putting in the limits we have
tan−1(y(x))− tan−1(1) = − ln cos x.
Since tan−1 = π/4 we have
y(x) = tan
(π
4
− ln cosx
)
.
(iii) Separating the variables in dx/dt = t2x we have
dx
x
= t2 dt,
and integrating gives
ln |x| = t
3
3
+ c.
Exponentiating both sides we have
|x(t)| = Aet3/3,
where A = ec is positive. Taking |x(t)| = x(t) gives a positive solu-
tion, while taking |x(t)| = −x(t) gives a negative solution. Thus the
general solution is x(t) = Aet
3/3, allowing any A ∈ R.
46 8 Separable equations
(iv) By separating the variables in dx/dt = −x2 we obtain the equation
−dx
x2
= dt,
and integrating both sides we obtain
1
x
= t + c.
It follows that
x(t) =
1
t + c
.
For the initial value problem with x(0) = x0, we need c = 1/x0,
and so the solution is
x(t) =
1
t + x−10
.
If x0 > 0 then x(t) → 0 as t →∞; if x0 < 0 then the solution blows
up in a finite time x(t) → −∞ as t → −x−10 .
(v) Separate variables in dy/dt = e−t2y2 to give
dy
y2
= e−t
2
dt,
and integrate,
∫ y(t)
y0
dy
y2
=
∫ t
0
e−t̃
2
dt̃.
Although we can integrate the left-hand side, we can find no closed
form integral for the right-hand side, so we leave it as it is:
[
−1
y
]y(t)
y=y0
=
∫ t
0
e−s
2
ds
(changing the dummy variable t̃ to an s). So
− 1
y(t)
+
1
y0
=
∫ t
0
e−s
2
ds,
which can be rearranged to give
y(t) =
1
y−10 −
∫ t
0 e
−s2 ds
.
Now, note that ∫ t
0
e−s
2
ds
Separable equations 47
is increasing in t, and tends to
√
π/2 as t → ∞. It follows that if
y0 < 2/
√
π then the solution is always finite, and reaches a limiting
value,
y(t) → 1
y−10 − (
√
π/2)
.
If y0 = 2/
√
π then as t → ∞, y(t) → ∞, while if y0 > 2/
√
π the
solution blows up in a finite time given by that value of t∗ for which
∫ t∗
0
e−s
2
ds = y−10 .
Exercise 8.2 Solve the linear equation
ẋ + px = q
by separation of variables.
Separating variables we have
1
q − px dx = dt,
and so, integrating
−1
p
ln |q − px| = t + c.
Multiplying up by −p and exponentiating both sides gives
|q − px| = Ae−pt
where A = e−pc. Taking either sign for the modulus gives positive or negative
values of A, and so we have
q − px(t) = Ae−pt,
and finally
x(t) = Be−pt +
q
p
(where B = −A/p could be any constant).
Exercise 8.3 Find the general solution of the equation
xy′ = ky
that is valid for x > 0.
48 8 Separable equations
After separating the variables the equation is
dy
y
=
k dx
x
.
Integrating both sides gives
ln |y| = k lnx + c,
since x > 0. Exponentiating this yields
|y| = Axk
with A = ec > 0. Depending on the sign of y(1) we obtain
y(x) = Axk
for any A ∈ R (since y(x) = 0 is clearly a solution).
Exercise 8.4 Find the function I(t) that satisfies
dI
dt
= p(t)I.
(Your answer will involve an integral.)
Separating the variables we obtain
dI
I
= p(t) dt.
Integrating both sides gives
ln |I| =
∫
p(t) dt
where we have used the notation
∫
p(t) dt to denote any anti-derivative of
p. Thus
|I(t)| = e
∫
p(t) dt,
or
I(t) = ±e
∫
p(t) dt.
Exercise 8.5 Use the method of separation of variables to show that the
general solution of the linear equation
ẋ = λx
is x(t) = Aeλt for any A ∈ R.
Separable equations 49
First notice that x(t) = 0 for all t is a solution. Now write
dx
x
= λdt,
an integrate both sides to give
ln |x| = λt + c.
Exponentiating we have
|x(t)| = Aeλt,
where A = ec > 0. Now, depending on the sign of x(t) we either have
x(t) = Aeλt with A > 0, or x(t) = Aeλt with A < 0. Thus the general
solution is
x(t) = Aeλt
for any A ∈ R.
Exercise 8.6 In Exercise 5.6 we showed, neglecting air resistance, that an
apple falling from a height h reaches the ground when t =
√
2h/g. If we
include air resistance then provided that v ≤ 0 the equation becomes
m
dv
dt
= −mg + kv2 v(0) = 0 (S8.1)
with k > 0. Show that
v(t) = −
√
mg
k
tanh
(√
gk
m
t
)
,
and hence that the apple now takes a time
t∗ =
√
m
kg
ln
(
ekh/m −
√
e2kh/m − 1
)
to reach the ground. Check that this coincides with the answer with no air
resistance (t∗ =
√
2h/g) as k → 0. Hint: for small x, ex ≈ 1 + x and
ln(1 + x) ≈ x.
Separating variables in (S8.1) we have
dv
κv2 − g = dt v(0) = 0,
where we have set κ = k/m for notational simplicity. Integrating both sides
of the differential equation between limits corresponding to times zero and
50 8 Separable equations
t gives
1
κ
∫ v
0
dṽ
ṽ2 − (g/κ) =
∫ t
0
dt̃,
which, using the standard integral
∫
1
a2 − x2 =
1
a
tanh−1
x
a
yields
−
[
1√
gκ
tanh−1
(
ṽ√
g/κ
)]v
ṽ=0
= t,
i.e.
− 1√
gκ
tanh−1
(
v√
g/κ)
= t,
which implies that
v(t) = −
√
g
κ
tanh(
√
gκ t).
Writing y(t) for the height above the ground, we now have
dy
dt
= −
√
g
κ
tanh(
√
gκ t) with y(0) = h.
Integrating both sides of the differential equation between times zero and t
gives
y(t) = h−
√
g
κ
∫ t
0
tanh(
√
gκ t̃) dt̃
= h−
√
g
κ
[
1√
gκ
ln cosh(
√
gκ t̃)
]t
t̃=0
= h− 1
κ
ln cosh(
√
gκ t).
since cosh 0 = 1.
The apple will hit the ground when y = 0, so when
ln cosh(
√
gκ t∗) = κh,
i.e. when
t∗ =
1√
gκ
cosh−1(eκh).
Separable equations 51
Since
coshx =
ex + e−x
2
,
it follows that if coshx = y then
e2x − 2yex + 1 = 0,
and solving this quadratic for ex gives
ex =
2y ±
√
4y2 − 4
2
= y ±
√
y2 − 1
and so
x = ln(y ±
√
y2 − 1).
We want the solution with t∗ > 0, so we take
t∗ =
1√
gκ
ln(eκh +
√
e2κh − 1),
as given in the question.
If k is small then κ is small, so we can use the approximations
ex ≈ 1 + x and ln(1 + x) ≈ x
(valid for small x) to write
t∗ ≈ 1√
gκ
ln(1 + κh +
√
2κh)
≈ 1√
gκ
ln(1 +
√
2κh),
≈ 1√
gκ
√
2κh
=
√
2h
g
,
where we have used the fact that when x is small,
√
x is much larger than
x.
Exercise 8.7 Show that for k 6= 0 the solution of the differential equation
dx
dt
= kx− x2 with x(0) = x0
is
x(t) =
k ektx0
x0(ekt − 1) + k .
52 8 Separable equations
Using this explicit solution describe the behaviour of x(t) as t →∞ for k < 0
and k > 0. (Note that this is much easier to do using the phase diagram
than using the explicit form of the solution.) For k = 0 see part (iv) of
Exercise 8.1.
Separating variables we have
dx
kx− x2 = dt.
Using partial fractions we have
1
k
[
1
x
+
1
k − x
]
dx = dt,
and so integrating both sides between the limits corresponds to times zero
and t we have ∫ x(t)
x0
1
x
+
1
k − x dx =
∫ t
0
k dt̃.
This gives
[
ln |x| − ln |k − x|
]x(t)
x=x0
= kt,
which is
ln |x(t)| − ln |k − x(t)| − ln |x0|+ ln |k − x0| = kt.
Exponentiating both sides gives
|x(t)||k − x0|
|x0||k − x(t)| = e
kt.
From the phase diagram, shown in Figure 8.1 (cf. Figure 7.14 above), we
can see that the signs of both x(t) and k − x(t) do not change over time.
k 0 0 k
Fig. 8.1. The phase diagram for the equation ẋ = kx− x2, for k < 0 (left) and for
k > 0 (right).
We can remove the modulus signs and then rearrange to get
x(t) =
k ektx0
x0(ekt − 1) + k .
Separable equations 53
When k > 0 the term x0ekt on the bottom will dominate, and so the solution
will tend to k. When k < 0 the denominator tends to k − 1, and the top
tends to zero, so x(t) → 0.
Exercise 8.8 Show that the solution of the equation
dx
dt
= −x(κ2 + x2)
with initial condition x(0) = x0 is
x(t) = ±
√
κ2
(1 + κ2x−20 )e2κ
2t − 1 ,
where the ± is chosen according to the sign of the initial condition. Deduce
that x(t) → 0 as t → ∞. As t decreases from zero the solution blows up as
t approaches a finite value t∗ < 0. When is this ‘blow up time’?
After separating variables we are left with
dx
x(κ2 + x2)
= −dt.
Using the method of partial fractions we can write
1
x(κ2 + x2)
=
1
κ2
(
1
x
− x
κ2 + x2
)
,
and so ∫ x(t)
x0
(
1
x
− x
κ2 + x2
)
dx = −
∫ t
0
κ2 dt̃,
which gives
[
ln |x| − 1
2
ln(κ2 + x2)
]x(t)
x=x0
= −κ2t,
or
ln
|x(t)|
√
κ2 + x20
|x0|
√
κ2 + x(t)2
= −κ2t.
Now we exponentiate both sides and square to give
x2(t)(κ2 + x20)
x20(κ2 + x(t)2)
= e−2κ
2t,
and after a rearrangement we can obtain
x(t)2 =
κ2
(1 + κ2x−20 )e2κ
2t − 1 .
54 8 Separable equations
Since ẋ < 0 for x > 0, and ẋ > 0 for x < 0, while ẋ = 0 if x = 0, it is clear
that the sign of x(t) does not change.
Thus depending on the sign of x0, we have
x(t) = ±
√
κ2
(1 + κ2x−20 )e2κ
2t − 1 .
As t →∞ the denominator tends to infinity, and so x(t) → 0. As t decreases
from zero the denominator decreases, and the solution will blow up as t
approaches t∗ where
(1 + κ2x−20 )e
2κ2t∗ − 1 = 0,
i.e.
t∗ = − 1
2κ2
ln(1 + κ2x−20 ).
Exercise 8.9 We found the solution of the equation ẋ = x(κ2 − x2) in
Section 8.6.1,
x(t) = ±
√
κ2
1 + e−2κ2t(κ2x−20 − 1)
.
Show that if |x0| > κ the solution blows up as t decreases towards a finite
negative value, and find this critical time.
The solution will blow up if the expression
1 + e−2κ
2t(κ2x−20 − 1)
becomes zero. If |x0| > κ then 0 < κ2x−20 < 1, and the expression κ2x−20 − 1
is negative. As t decreases from zero the contribution of the second term
becomes larger, and will tends to −1 as t → t∗ with
1 + e−2κ
2t∗(κ2x−20 − 1) = 0,
so
t∗ =
1
2κ2
ln(1− κ2x−20 ).
[This is negative, since 1− κ2x−20 < 1.]
Exercise 8.10 Consider the equation
ẋ = xα with x(0) ≥ 0
for α > 0. Show that the only value of α for which the equation has solutions
that are both unique and exist for all time is α = 1. You should be able to
Separable equations 55
find an initial condition for which the solutions are not unique when α < 1
(cf. (6.3)), and show that solutions with x(0) > 0 blow up in a finite time if
α > 1 (cf. (6.6)).
First we solve the equation by separating the variables,
dx
xα
= dt.
Integrating from times zero to t gives
[
1
1 + α
x1+α
]x(t)
x0
= t,
which implies that
x(t)1−α − x1−α0 = (1− α)t,
and so
x(t) = [(1− α)t + x1−α0 ]1/(1−α).
If α < 1 then (cf. example ẋ = x1/2 in section 6.1) for any choice of c ≥ 0
the function
x(t) =
{
0 t ≤ 0
[(1− α)(t− c) + x1−α0 ]1/(1−α) t ≥ c
satisfies ẋ = xα with x(0) = 0.
If α > 1 then for positive initial conditions to solution will blow up in
finite time, since 1− α < 0: x(t) → +∞ as t → t∗, where
t∗ =
x1−α0
α− 1
(cf. the example ẋ = x2 in section 6.3).
If α = 1 then solutions are unique since f(x) = x and f ′(x) = 1 are both
continuous for all x ∈ R, and the explicit form of the solution x(t) = x0et
shows that the solution exists for all time.
Exercise 8.11 Assuming that f(x) and f ′(x) are continuous, show that if
the solution of
ẋ = f(x) with x(0) = x0
blows up to x = +∞ in finite time then
∫ ∞
x0
f(x) dx < ∞.
56 8 Separable equations
The solution of the equation can be found by separating variables:
dx
f(x)
= dt.
Integrating between times zero and t gives
∫ x(t)
x0
f(x) dx = t. (S8.2)
If there is a solution with x(0) = x0 that blows up to x = +∞ in a finite
time, i.e. as t → t∗, then taking limits as t → t∗ on both sides of (S8.2) we
get ∫ ∞
x0
f(x) dx = t∗ < ∞.
9
First order linear equations and the integrating
factor
Exercise 9.1 Use an integrating factor to solve the following differential
equations:
(i)
dy
dx
+
y
x
= x2
(find the general solution and the only solution that is finite when
x = 0),
(ii)
dx
dt
+ tx = 4t
(find the solution with x(0) = 2),
(iii)
dz
dy
= z tan y + sin y
(find the general solution),
(iv)
y′ + e−xy = 1
(find the solution when y(0) = e, leaving your answer as an integral),
(v)
ẋ + x tanh t = 3
(find the general solution, and compare it to that for ẋ + x = 3),
(vi)
y′ + 2y cotx = 5
(find the solution with y(π/2) = 1),
57
58 9 First order linear equations and the integrating factor
(vii)
dx
dt
+ 5x = t
(find the general solution),
(viii) with a > 0 find the solution of the equation
dx
dt
+
[
a +
1
t
]
x = b
for a general initial condition x(1) = x0, and show that x(t) → b/a
as t → ∞ (you would get the same result if you replaced a + t−1 by
a).
(i) The integrating factor for
dy
dx
+
y
x
= x2
is
I(x) = exp
(∫
1
x
dx
)
= exp lnx = x,
and so we have
x
dy
dx
+ y =
d
dx
(xy) = x3.
Integrating both sides with respect to x gives
xy =
x4
4
+ c
and so
y(x) =
x3
4
+
c
x
.
Note that we only have a solution near x = 0 if y(0) = 0, and this
solution is y(x) = x3/4.
(ii) The integrating factor for
dx
dt
+ tx = 4t
is
I(t) = exp
(∫
t dt
)
= et
2/2.
Therefore we have
et
2/2 dx
dt
+ et
2/2tx =
d
dt
(xet
2/2) = 4tet
2/2.
First order linear equations and the integrating factor 59
Integrating both sides between 0 and t gives
x(t)et
2/2 − x(0) = 4et2/2 − 4
or, since x(0) = 2
x(t) = 4− 2e−t2/2.
(iii) Before looking for the integrating factor we need to rearrange the
equation
dz
dy
=z tan y + sin y
into the standard form
dz
dy
− (tan y)z = sin y.
Now the integrating factor is
I(y) = exp
(∫
− tan y dy
)
= exp(ln cos y) = cos y,
and so we have
cos y
d
dy
+ z sin y =
d
dy
(z cos y) = sin y cos y = 12 sin 2y.
Integrating both sides gives
z(y) cos y = −cos 2y
4
+ c,
and so
z(y) = − cos 2y
4 cos y
+
c
cos y
.
(iv) The integrating factor for the equation
y′ + e−xy = 1
is
I(x) = exp
(∫
e−x
)
= exp(−e−x) = e−e−x .
Then we have
d
dx
(
ye−e
−x)
= e−e
−x
,
and integrating between zero and x we get
y(x)e−e
−x − y(0)e−e0 =
∫ x
0
e−e
−x̃
dx̃,
60 9 First order linear equations and the integrating factor
which simplifies, since y(0) = e, to give
y(x) = 1 + ee
−x
∫ x
0
e−e
−x̃
dx̃.
(v)
ẋ + x tanh t = 3.
Here the integrating factor is
I(t) = exp
(∫
tanh t dt
)
= exp(ln cosh t) = cosh t.
So we have
d
dt
(x cosh t) = 3 cosh t.
Integrating both sides gives
x(t) cosh t = 3 sinh t + c,
and so
x(t) = 3 tanh t + c sech t.
As t →∞,
tanh t =
et − e−t
et + e−t
=
1− e−2t
1 + e−2t
≈ 1− 2e−2t
and
sech t =
2
et + e−t
≈ 2e−t.
So as t →∞ we have x(t) ≈ 3 + 2ce−t.
To find the solution of ẋ + x = 3 we use the integrating factor et,
and so
d
dt
(xet) = 3et ⇒ x(t)et = 3et + d ⇒ x(t) = 3 + de−t.
The equations have similar solutions for t large; this is not surprising,
since for large t we have tanh t ≈ 1 (as shown above).
(vi)
y′ + 2y cotx = 5.
The integrating factor is
I(x) = exp
(∫
2 cot xdx
)
= exp(2 ln | sinx|) = | sinx|2 = sin2 x.
First order linear equations and the integrating factor 61
Then we have
d
dx
(y sin2 x) = 5 sin2 x.
Integrating between π/2 and x,
[
y(x̃) sin2 x̃
]x
x̃=π/2
= 5
∫ x
π/2
sin2 x̃dx̃
we obtain
y(x) sin2 x− 1 = 5
∫ x
π/2
1− cos 2x̃
2
dx̃
= 5
[
x
2
− sin 2x
4
]x
x̃=π/2
= 5
(
x
2
− sin 2x
4
− π
4
)
,
and so
y(x) =
1
sin2 x
[
1 + 5
(
x
2
− sin 2x
4
− π
4
)]
.
(vii)
dx
dt
+ 5x = t.
The integrating factor is e5t, and so
d
dt
(e5tx(t)) = te5t.
Integrating both sides gives
e5tx(t) =
∫
te5t dt
=
te5t
5
− 1
5
∫
e5t dt
=
te5t
5
− e
5t
25
+ c,
and so
x(t) =
t
5
− 1
25
+ ce−5t.
(viii) The integrating factor for
dx
dt
+
[
a +
1
t
]
x = b
62 9 First order linear equations and the integrating factor
is
exp
(∫
a +
1
t
dt
)
= exp(at + ln t) = teat.
Multiplying both sides by this we get
d
dt
(x(t)teat) = bteat.
Integrating between times 1 and t gives
x(t)teat − x0ea =
∫ t
1
bt̃eat̃ dt̃.
Integrating the right-hand side by parts gives
x(t)teat − x0ea =
[
bt̃eat̃
a
]t
t̃=1
−
∫ t
1
beat̃
a
dt̃
=
bteat − bea
a
−
[
beat̃
a2
]t
t̃=1
= b
(
teat − ea
a
− e
at − ea
a2
)
.
Rearranging gives
x(t) =
b
a
+
(
x0 +
b
a
+
b
a2
)
ea(1−t) − b
a2t
,
and so x(t) → b/a as claimed.
Exercise 9.2 A body is found in a cold room (temperature 5◦C) at 3 p.m.
and its temperature then is 19◦C. An hour later its temperature has dropped
to 15◦C. Use Newton’s law of cooling to estimate the time of death, assuming
that body temperature is 37◦C.
Newton’s law of cooling is
dT
dt
= −k(T −A(t)) or dT
dt
+ kT = kA(t).
When A(t) = A, a constant, we can use the integrating factor ekt and then
we have
d
dt
(ektT ) = kAekt.
Integrating between t0 and t we obtain
ektT (t)− ekt0T (t0) = A(ekt − ekt0),
First order linear equations and the integrating factor 63
or
T (t) = A + e−k(t−t0)[T (t0)−A]. (S9.1)
First we find k using T (3) = 19, T (4) = 15, and A = 5:
15 = T (4) = 5 + e−k[19− 5] ⇒ 10 = 14e−k
which gives k = − ln(5/7) ≈ 0.337.
If the time of death was t0 then, using T (t0) = 37, we have
19 = T (3) = 5 + e−k(3−t0)[37− 5] ⇒ 14 = 32e−k(3−t0),
so that
t0 = 3 +
ln(14/32)
k
≈ 0.543,
which fixes the time of death at approximately 12:33 a.m.
Exercise 9.3 At 7 a.m. in the morning I make my wife a cup of tea using
boiling water; after adding some milk it is about 90◦C. When we leave for
the station at 7:30 a.m. the tea is still drinkable at about 45◦C. When I get
back home at 8 a.m. the neglected tea has cooled to about 30◦C. What is the
temperature of our house?
We use the solution (S9.1) from the previous exercise. We have T (7) = 90,
T (7.5) = 45, and T (8) = 30. Therefore
45 = A + e−k/2[90−A] and 30 = A + e−k[90−A].
Since e−k/2 = (45−A)/(90−A) we have
30 = A +
(45−A)2
90−A ,
which gives an equation for A,
(30−A)(90−A) = (45−A)2.
Solving this equation gives A = 22.5, and so the house is 22.5◦C.
Exercise 9.4 Use the integrating factor method to find T (t2) in terms of
T (t1) when
dT
dt
= −k(T (t)−A(t))
and
A(t) = µ + a cosω(t− φ).
64 9 First order linear equations and the integrating factor
Rearrange the equation as
dT
dt
+ kT = kA(t).
The integrating factor is
exp
(∫
k dt
)
= ekt,
and so we have
d
dt
(T (t)ekt) = kektA(t).
Integrating both sides between t1 and t2 gives
T (t2)ekt2 − T (t1)ekt1 = k
∫ t2
t1
ektA(t) dt.
With the choice A(t) = µ + a cosω(t− φ) we get
T (t2)ekt2 − T (t1)ekt1 = kµ
∫ t2
t1
ekt dt + ak
∫ t2
t1
ekt cosω(t− φ) dt.
An anti-derivative of ekt cosω(t− φ) is (cf. example in Section 9.4.2)
k
k2 + ω2
ekt cosω(t− φ) + ω
k2 + ω2
ekt sinω(t− φ),
and so
T (t2)ekt2 − T (t1)ekt1 = µ[ekt2 − ekt1 ]
+ak
[
k
k2 + ω2
ekt2 cosω(t2 − φ) + ω
k2 + ω
ekt2 sinω(t2 − φ)
]
−ak
[
k
k2 + ω2
ekt1 cosω(t1 − φ) + ω
k2 + ω
ekt1 sinω(t1 − φ)
]
.
Rearranging we have
T (t2) = µ + ak
[
k
k2 + ω2
cosω(t2 − φ) + ω
k2 + ω
sinω(t2 − φ)
]
+
[
T (t1)− µ− ak
k2 + ω2
(k cosω(t1 − φ) + ω sinω(t1 − φ))
]
e−k(t2−t1).
Exercise 9.5 A dead body is found outside on a winter’s morning at 7 a.m.;
its temperature is measured as 20◦C. Measured an hour later it has dropped
to 15◦C. The air temperature A(t) fluctuates on a daily cycle about a mean
of 3◦C with A(t) = 3−5 cos ω(t−2), where t is measured in hours with t = 0
corresponding to midnight, and ω = π/12.
First order linear equations and the integrating factor 65
(i) Use the solution from question 9.4 and the temperature observations
at 7 a.m. and 8 a.m. to show that
k = − ln
{
12(k2 + ω2)− 5k(k cos 6ω + ω sin 6ω)
17(k2 + ω2)− 5k(k cos 5ω + ω sin 5ω)
}
. (S9.2)
(ii) (C) This is a MATLAB exercise. Choose an initial guess for k, and
then substitute this into the right-hand side of (S9.2) to obtain a new
guess. Continue doing this until your ‘guess’ stabilises. Once this
happens you have actually obtained the required solution of (S9.2).
Can you see why? [You should find k ≈ 0.3640.]
(iii) If the time of death was t0, use the fact that body temperature is 37◦C
(so T (t0) = 37) and T (7) = 20 to show that
t0 = 7 +
1
k
ln
{
17(k2 + ω2)− 5k(k cos 5ω + ω sin 5ω)
34(k2 + ω2)− 5k(k cosω(t0 − 2) + ω sinω(t0 − 2))
}
.
(iv) (C) Use MATLAB again to refine an initial guess for the time of death
as in part (ii). You should find that t0 ≈ 4.8803, or 4:53 a.m.
(i) We have T (7) = 20, T (8) = 15, µ = 3, a = 5, φ = 2, and ω = π/12.
We need to find t0 such that T (t0) = 37. We should be able to use
the information from 7 a.m. and 8 a.m. to estimate k. Our solution
gives (leaving ω = π/12 in the equations)
15 = 3 + 5k
[
k
k2 + ω2
cos 6ω +
ω
k2 + ω2
sin 6ω
]
+
[
17− 5k
k2 + ω2
(k cos 5ω + ω sin 5ω)
]
e−k.
No matter how we rearrange this equation we cannot solve it explic-
itly for k. However, if we rearrange it as
e−k =
12(k2 + ω2)− 5k(k cos 6ω + ω sin 6ω)
17(k2 + ω2)− 5k(k cos 5ω + ω sin 5ω)
we can obtain an equation for k ‘in terms of k’:
k = − ln
{
12(k2 + ω2)− 5k(k cos 6ω + ω sin 6ω)
17(k2 + ω2)− 5k(k cos 5ω + ω sin 5ω)
}
. (S9.3)
(ii) What we can do with this is to guess k, and then substitute our guess
into the right-hand side to obtain a new guess for k, and continue like
this until, if we are lucky, our guess settles down to a fixed value which
will be the solution we want.
66 9 First order linear equations and the integrating factor
Writing this more mathematically, given an initial guess k0 we then
calculate successive guesses kn by setting
kn+1 = − ln
{
12(k2n + ω
2)− 5kn(kn cos 6ω + ω sin 6ω)
17(k2n + ω2)− 5kn(kn cos 5ω + ω sin 5ω)
}
.
If kn+1 =k + n = k then k must be a solution of (S9.3).
You can do this fairly easily in MATLAB, and the result is k ≈
0.3640. [The M-file findk.m will do this for you; you can check your
answer with the file temperature.m which will compute the solution
at time t2 given the temperature at time t1 and the value of k.]
(iii) To find the time of death will require a similar computer calculation,
since once again we have an equation that we cannot solve explicitly.
If we set T (t0) = 37 and T (7) = 20 then we should be able to find t0
from (we don’t replace k by 0.3640 here to simplify the algebra)
20 = 3 + 5k
[
k
k2 + ω2
cos 5ω +
ω
k2 + ω2
sin 5ω
]
+
[
34− 5k
k2 + ω2
(k cosω(t0 − 2) + ω sinω(t0 − 2))
]
e−k(7−t0)
We can rearrange this as
e−k(7−t0) =
17(k2 + ω2)− 5k(k cos 5ω + ω sin 5ω)
34(k2 + ω2)− 5k(k cosω(t0 − 2) + ω sinω(t0 − 2)) ,
or
t0 = 7+
1
k
log
{
17(k2 + ω2)− 5k(k cos 5ω + ω sin 5ω)
34(k2 + ω2)− 5k(k cosω(t0 − 2) + ω sinω(t0 − 2))
}
.
(iv) Iterating this (you could use the M-file findt0.m) gives t0 ≈ 4.8803,
which puts the time of death at 4:53 a.m. (Again, you could check
that this works with the temperature.m file.)
Exercise 9.6 Show that if y1 and y2 are any two solutions of
dy
dx
+ p(x)y = 0
then y1(x)/y2(x) is constant. (You do not need to solve the equation!)
Differentiating y1(x)/y2(x) with respect to x gives
d
dx
[
y1(x)
y2(x)
]
=
y2y
′
1 − y1y′2
y22
=
y2[−p(x)y1]− y1[−p(x)y2]
y22
First order linear equations and the integrating factor 67
= 0,
and so y1(x)/y2(x) is constant.
Exercise 9.7 Suppose that
dx
dt
≤ ax
(this is known as a differential inequality). Use an appropriate integrating
factor to show that
d
dt
[
e−atx
] ≤ 0,
and then integrate both sides between appropriate limits to deduce that
x(t) ≤ x(s)ea(t−s)
for any t and s. Hint: it is a fundamental property of integration that if
f(x) ≤ g(x) then
∫ b
a
f(x) dx ≤
∫ b
a
g(x) dx.
We can rewrite the equation as
dx
dt
− ax ≤ 0,
and then use the integrating factor eat to obtain
eat
[
dx
dt
− ax
]
≤ 0
(this is allowed since eat is always positive). The left-hand side of the equa-
tion is just ddt(e
−atx), so we now have
d
dt
[
e−atx
] ≤ 0
as required. Integrating both sides between times s and t gives
e−atx(t)− e−asx(s) ≤ 0
which on rearrangement yields
x(t) ≤ e−a(t−s)x(s).
68 9 First order linear equations and the integrating factor
Exercise 9.8 The function sinωt can be written as a combination of complex
exponentials,
sinωt =
eiωt − e−iωt
2i
.
Using this form for sinωt, and assuming that the usual rules of integration
apply to such complex exponentials, find
∫
ekt sinωt dt.
You may also need to use the identity
cosωt =
eiωt + e−iωt
2
.
See Appendix A for more on these complex exponentials.
We write
∫
ekt sinωt dt =
∫
ekt
eiωt − e−iωt
2i
dt
=
1
2i
∫
e(k+iω)t − ek−iω)t dt
=
1
2i
[
e(k+iω)t
k + iω
− e
(k−iω)t
k − iω
]
=
ekt
2i
[
eiωt
k + iω
− e
−iωt
k − iω
]
= ekt Im [(k − iω)(cosωt + i sinωt)]
= ekt(k sinωt− ω cosωt)
since z − z∗ = 2i Im[z].
10
Two ‘tricks’ for nonlinear equations
Exercise 10.1 Check that the following equations are exact and hence solve
them.
(i)
(2xy − sec2 x) + (x2 + 2y)dy
dx
= 0,
(ii)
(1 + exy + xexy) + (xex + 2)
dy
dx
= 0,
(iii)
(x cos y + cosx)
dy
dx
+ sin y − y sinx = 0,
and
(iv)
ex sin y + y + (ex cos y + x + ey)
dy
dx
= 0.
(i) We have
(2xy − sec2 x)︸ ︷︷ ︸
f(x,y)
+ (x2 + 2y)︸ ︷︷ ︸
g(x,y)
dy
dx
= 0.
To check that this equation this exact, i.e. that there is an F (x, y)
such that
∂F
∂x
= f(x, y) and
∂F
∂y
= g(x, y)
we need to make sure that
∂f
∂y
=
∂g
∂x
.
69
70 10 Two ‘tricks’ for nonlinear equations
So we check
∂f
∂y
= 2x =
∂g
∂x
.
Now to find F we first solve
∂F
∂x
= 2xy − sec2 x.
Integrating with respect to x we get
F (x, y) = x2y − tanx + C(y).
In order to find C(y) we partially differentiate F (x, y) with respect
to y,
∂F
∂y
= x2 +
dC
dy
= G(x, y) = x2 + 2y,
and so
dC
dy
= 2y
which gives C(y) = y2 + c. So the solution is
F (x, y) = x2y − tanx + y2 = c.
(ii) Now for
(1 + exy + xexy)︸ ︷︷ ︸
f(x,y)
+ (xex + 2)︸ ︷︷ ︸
g(x,y)
dy
dx
= 0
we check
∂f
∂y
= ex + xex =
∂g
∂x
and the equation is exact. First we solve
∂F
∂x
= 1 + exy + xexy,
integrating both sides with respect to x to give
F (x, y) = x + xexy + C(y).
So we have
∂F
∂y
= xex +
dC
dy
= xex + 2,
so in fact
dC
dy
= 2
Two ‘tricks’ for nonlinear equations 71
which gives C = 2y. Therefore F (x, y) = x + xexy + 2y, and we have
x + xexy + 2y = C.
This simplifies to give
y =
C − x
2 + xex
.
(iii) For
(sin y − y sinx)︸ ︷︷ ︸
f(x,y)
+(x cos y + cosx)︸ ︷︷ ︸
g(x,y)
dy
dx
= 0
we check that
∂f
∂y
= cos y − sinx = ∂g
∂x
,
and so the equation is exact.
Now we solve
∂F
∂x
= sin y − y sinx,
which gives
F (x, y) = x sin y + y cosx + C(y).
In order to find C(y) we differentiate partially with respect to y, and
then
∂F
∂y
= x cos y + cosx + C ′(y) = g(x, y) = x cos y + cosx,
and so C ′(y) = 0 and C is a constant. So the solution is
x sinx + y cosx = c
for any constant c.
(iv) Now we have
(ex sin y + y)︸ ︷︷ ︸
f(x,y)
+(ex cos y + x + ey)︸ ︷︷ ︸
g(x,y)
dy
dx
= 0,
and we check that it is exact by showing that
∂f
∂y
= ex cos y + 1 =
∂g
∂x
.
We first solve the equation
∂F
∂x
= ex sin y + y,
72 10 Two ‘tricks’ for nonlinear equations
so that
F (x, y) = ex sin y + xy + C(y).
Now we have
ex cos y + x + ey =
∂F
∂y
= ex cos y + x + C ′(y).
and so C ′(y) = ey which implies that C(y) = ey + c. So we obtain
the solution
ex sin y + xy + ey + c = 0..
Exercise 10.2 Find an integrating factor depending only on x that makes
the equation
e−y secx + 2 cotx− e−y dy
dx
= 0
exact, and hence find its solution. Hint:
∫
cosecxdx = ln |cosec x− cotx|.
To look for an integrating factor I(x) depending only on x we want
I(x)(e−y secx + 2 cotx)︸ ︷︷ ︸
f(x,y)
+ (−I(x)e−y)︸ ︷︷ ︸
g(x,y)
dy
dx
= 0
to be exact. This needs
∂f
∂y
= −I(x)e−y secx = −I ′(x)e−y = ∂g
∂x
,
i.e.
I ′(x) = (secx)I(x).
The solution of this is I(x) = e
∫
sec(x) dx = eln(sec x+tan x) = secx + tanx.
Using this integrating factor the original equation becomes
(sec x + tanx)(e−y sec x + 2 cotx)︸ ︷︷ ︸
f̃(x,y)
+ [−(sec x + tanx)e−y]︸ ︷︷ ︸
g̃(x,y)
dy
dx
= 0.
This new equation is exact, since
∂f̃
∂y
= (sec x + tanx)(−e−y secx) = −e−y(sec2 x + secx tanx) = ∂g̃
∂x
.
So we can now find F (x, y): we will do this by solving
∂F
∂y
= g̃(x, y) = −(secx + tanx)e−y,
Two ‘tricks’ for nonlinear equations 73
giving
F (x, y) = e−y(sec x + tanx) + C(x).
Now we want ∂F∂x = f̃(x, y), so we need
e−y(secx tanx+sec2 x)+C ′(x) = e−y secx(sec x+tanx)+2 cotx(sec x+tanx),
which gives
C ′(x) = 2 cosecx + 2 ⇒ 2 ln |cosecx− cotx|+ 2x + c,
and so
e−y(sec x + tanx) + 2 ln |cosec x− cotx|+ 2x + c = 0.
Exercise 10.3 Show that any equation that can be written in the form
f(x) + g(y)
dy
dx
= 0
is exact, and find its solution in terms of integrals of f and g. Hence find
the solutions of
(i)
V ′(x) + 2y
dy
dx
= 0
and
(ii) (
1
y
− a
)
dy
dx
+
2
x
− b = 0,
for x, y > 0.
Here we have
∂f
∂y
= 0 =
∂g
∂y
.
First we can solve
∂F
∂x
= f(x)
giving
F (x, y) =
∫
f(x) dx + C(y).
We then need
∂F
∂y
= C ′(y) = g(y),
74 10 Two ‘tricks’ for nonlinear equations
and so C(y) =
∫
g(y) dy + c, giving the solution
∫
f(x) dx +
∫
g(y) dy + c = 0.
(i) The solution is V (x) + y2 = E.
(ii) The solution is ln y − ay + 2 lnx− bx = c.
Exercise 10.4 By substituting u = y/x solve the following homogeneous
equations:
(i)
xy + y2 + x2 − x2 dy
dx
= 0
(the solution is y = x tan(ln |x|+ c)).
(ii)
dx
dt
=
x2 + t
√
t2 + x2
tx
(the solution is x(t) = ±t
√
(ln |t|+ c)2 − 1).
(i) Divide through by x2, then the equation becomes
dy
dx
=
y
x
+
y2
x2
+ 1.
This is clearly homogeneous. If we put u = y/x then
u′ = − y
x2
+
y′
x
,
and so
u′ = −u
x
+
u + u2 + 1
x
,
i.e.
x
du
dx
= 1 + u2.
Separating the variables and integrating we have
∫
du
1 + u2
du =
∫
1
x
dx,
and so
tan−1(u) = ln |x|+ c,
or u = tan(ln |x|+ c). Since y = xu weget
y(x) = x tan(ln |x|+ c).
Two ‘tricks’ for nonlinear equations 75
(ii) Rearranging the equation
dx
dt
=
x2 + t
√
t2 + x2
tx
we have
dx
dt
=
x
t
+
√
(t/x)2 + 1
which is clearly homogeneous. So we substitute u = x/t, and then
u̇ = − x
t2
+
ẋ
t
or
tu̇ = −u + ẋ,
which gives
tu̇ = −u + u +
√
u−2 + 1 =
√
u−2 + 1.
So separating variables we have
du√
u−2 + 1
=
1
t
dt
which gives after a rearrangement of the left-hand side
∫
udu√
1 + u2
=
∫
1
t
dt.
Therefore
(1 + u2)1/2 = ln |t|+ c.
Squaring both sides we have
(1 + u2) = (ln |t|+ c)2,
and so
u = ±
√
(ln |t|+ c)2 − 1,
which, since y = tu, gives the required solution.
Exercise 10.5 You could solve
dx
dt
= kx− x2.
by separating variables (see Exercise 8.7). Instead, substitute u = x−1 and
show that u satisfies the linear equation
du
dt
= 1− ku.
76 10 Two ‘tricks’ for nonlinear equations
Solve this equation for u(t), and hence find the solution x(t).
Making the substitution u = x−1 we have du/dt = −x−2ẋ, and so
du
dt
= − 1
x2
[kx− x2] = −k
x
+ 1 = 1− ku.
This is
du
dt
+ ku = 1
which we can solve with the integrating factor ekt,
d
dt
(uekt) = ekt.
Integrating both sides we get
u(t)ekt =
ekt
k
+ c,
and so
u(t) =
1
k
+ ce−kt.
Since x(t) = u(t)−1 we have
x(t) =
1
k−1 + ce−kt
=
k
1 + cke−kt
which is the same answer as before (Exercise 8.7) if we set A = ck.
Exercise 10.6 Use an appropriate substitution to solve the equation
ẋ = x(κ2 − x2).
You should recover the solution (8.16) found by separating variables.
This is a Bernoulli equation: we can put u = x−2 and then
u̇ = −2x−3ẋ
= −2κ2x−2 + 2
= 2− 2κ2u.
Solving the equation
du
dt
+ 2κ2u = 2
using the integrating factor e2κ
2t we obtain
d
dt
(ue2κ
2t) = 2e2κ
2t
Two ‘tricks’ for nonlinear equations 77
and so, integrating between times zero and t
u(t)e2κ
2t = u(0) +
e2κ
2t − 1
κ2
,
which gives
u(t) = u(0)e−2κ
2t +
1− e−2κ2t
κ2
.
Since u(t) = x(t)−2 we have
x(t)−2 = x−20 e
−2κ2t +
1− e−2κ2t
κ2
.
Rearranging this for x(t) gives
x(t) = ±
√
κ2
1 + e−2κ2t(κ2x−20 − 1)
,
the same solution we obtain using the method of separation of variables as
(8.16).
11
Second order linear equations: general theory
Exercise 11.1 By finding the Wronskian of the following pairs of solutions,
show that they are linearly independent:
(i) x1(t) = ek1t and x2(t) = ek2t with k1 6= k2,
(ii) x1(t) = ekt and x2(t) = tekt, and
(iii) x1(t) = eρt sinωt and x2(t) = eρt cosωt.
In each case we want to show that W (x1, x2) 6= 0. We have
(i)
W (ek1t, ek2t) = ek1tk2ek2t − ek2tk1ek1t = (k2 − k1)e(k1+k2)t;
(ii)
W (ekt, tekt) = ekt(ekt + ktekt)− tektkekt = e2kt;
and
(iii)
W (eρt sinωt, eρt cosωt)
= eρt sinωt(ρeρt cosωt
−ωeρt sinωt)− eρt cosωt(ρeρt sinωt + ωeρt cosωt)
= −ωe2ρt.
In each case W (x1, x2) 6= 0, and so the two solutions are linearly indepen-
dent.
Exercise 11.2 Show that the Wronskian for two solutions x1(t) and x2(t)
of the second order differential equation
d2x
dt2
+ p1(t)
dx
dt
+ p2(t)x = 0 (S11.1)
78
Second order linear equations: general theory 79
satisfies
Ẇ (t) = −p1(t)W (t).
(Write W (t) = x1(t)ẋ2(t) − x2(t)ẋ1(t), differentiate, and use the fact that
x1(t) and x2(t) satisfy the equation (S11.1).) Deduce that either W (t) = 0
for all t, or W (t) 6= 0 for all t.
We have W (t) = x1ẋ2 − x2ẋ1, and so
Ẇ = ẋ1ẋ2 + x1ẍ2 − ẋ2ẋ1 − x2ẍ1
= x1ẍ2 − x2ẍ1
= x1[−p1ẋ2 − p2x2]− x2[−p1ẋ1 − p2x1]
= −p1[x1ẋ2 − x2ẋ1]
= −p1W.
Solving this linear equation (use an integrating factor) for W gives
W (t) = W (0) exp
(∫ t
0
−p1(t̃) dt̃
)
.
Since ex is always positive, either W (t) is never zero (if W (0) 6= 0) or W (t)
is identically zero (if W (0) = 0).
Exercise 11.3 We have seen that if x1 and x2 are two solutions of a linear
differential equation, then they are linearly independent if and only if their
Wronskian is non-zero. The simple example of this question shows that this
is not true for general functions that are not the solutions of some differential
equation.
(i) Check carefully that if f(t) = t2|t| then df/dt = 3t|t| (this is easy
when t 6= 0; you will have to use the formal definition of the derivative
at t = 0).
(ii) Let
f1(t) = t2|t| and f2(t) = t3.
Show that although these two functions are linearly independent on
R, their Wronskian is identically zero.
(i) For t < 0, f(t) = −t3, so ḟ = −3t2 = 3t|t|; for t > 0 similarly
f(t) = t3 and ḟ = 3t2 = 3t|t|. At t = 0,
ḟ(0) = lim
h→0
f(h)− f(0)
h
= lim
h→0
h2|h|
h
= lim
h→0
h|h| = 0.
So ḟ = 3t|t| as claimed.
80 11 Second order linear equations: general theory
(ii) We have ḟ1(t) = 3t|t| from part (i), and ḟ2(t) = 3t2. Therefore
W [f1, f2](t) = [t2|t| × 3t2]− [t3 × 3t|t|] = 3t4|t| − 3t4|t| = 0.
However, it is clear that the functions are linearly independent on R,
since if
αt2|t|+ βt3 = 0 for all t ∈ R
then taking any t > 0 implies that α = −β, while taking any t < 0
gives α = β, and so the only solution is α = β = 0.
12
Homogeneous linear 2nd order equations with
constant coefficients
Exercise 12.1 Find the general solution of the following differential equa-
tions, and then the solution satisfying the specified initial conditions.
(i) ẍ− 3ẋ + 2x = 0 with x(0) = 2 and ẋ(0) = 6;
(ii) y′′ − 4y′ + 4y = 0 with y(0) = 0 and y′(0) = 3;
(iii) z′′ − 4z′ + 13z = 0 with z(0) = 7 and z′(0) = 42;
(iv) ÿ + ẏ − 6y = 0 with y(0) = −1 and ẏ(0) = 8;
(v) ÿ − 4ẏ = 0 with y(0) = 13 and ẏ(0) = 0;
(vi) θ̈ + 4θ = 0 with θ(0) = 0 and θ̇(0) = 10;
(vii) ÿ + 2ẏ + 10y = 0 with y(0) = 3 and ẏ(0) = 0;
(viii) 2z̈ + 7ż − 4z = 0 with z(0) = 0 and ż(0) = 9;
(ix) ÿ + 2ẏ + y = 0 with y(0) = 0 and ẏ(0) = −1;
(x) ẍ + 6ẋ + 10x = 0 with x(0) = 3 and ẋ(0) = 1;
(xi) 4ẍ− 20ẋ + 21x = 0 with x(0) = −4 and ẋ(0) = −12;
(xii) ÿ + ẏ − 2y = 0 with y(0) = 4 and ẏ(0) = −4;
(xiii) ÿ − 4y = 0 with y(0) = 10 and ẏ(0) = 0;
(xiv) y′′ + 4y′ + 4y = 0 with y(0) = 27 and y′(0) = −54; and
(xv) ÿ + ω2y = 0 with y(0) = 0 and ẏ(0) = 1.
(i)
ẍ− 3ẋ + 2x = 0 with x(0) = 2 and ẋ(0) = 6.
Try x = ekt and obtain the auxiliary equation for k,
k2 − 3k + 2 = 0
with solutions k = 2 and k = 4. The general solution is therefore
x(t) = Ae2t + Be4t.
81
82 12 Homogeneous 2nd order linear ODEs
Since ẋ(t) = 2Ae2t + 4Be4t the particular solution with
x(0) = A + B = 2 and ẋ(0) = 2A + 4B = 6
has A = B = 1 and is
x(t) = e2t + e4t.
(ii)
y′′ − 4y′ + 4y = 0 with y(0) = 0 and y′(0) = 3.
Try y(x) = ekx to obtain the auxiliary equation
k2 − 4k + 4 = 0
which has k = 4 as a repeated root. The general solution is therefore
y(x) = Ae4x + Bxe4x.
Since y′(x) = 4Ae4x + B[e4x + 4xe4x] the solution with
y(0) = A + B = 0 and y′(0) = 4A + B = 3
has A = 1 and B = −1: it is
y(x) = (1− x)e4x.
(iii)
z′′ − 4z′ + 13z = 0 with z(0) = 7 and z′(0) = 42.
Try z(x) = ekx to obtain
k2 − 4k + 13 = 0,
with solutions k = 4± 6i. The general solution is therefore
z(x) = e4x(A cos 6x + B sin 6x).
We have
z′(x) = e4x(4A cos 6x + 4B sin 6x− 6A sin 6x + 6B cos 6x)
= e4x[(4A + 6B) cos 6x + (4B − 6A) sin 6x],
and so the solution with
z(0) = A + B = 7 and z′(0) = 4A + 6B = 42
has A = 0 and B = 7, and is
z(x) = 7e4x sin 6x.
Homogeneous 2nd order linear ODEs 83
(iv)
ÿ + ẏ − 6y = 0 with y(0) = −1 and ẏ(0) = 8.
Trying y(t) = ekt gives the auxiliary equation
k2 + k − 6 = 0
with solutions k = 2 or k = −3. So the general solution is
y(t) = Ae2t + Be−3t.
Since ẏ(t) = 2Ae2t − 3Be−3t the solution with
y(0) = A + B = −1 and ẏ(0) = 2A− 3B = 8
has A = 1 and B = −2; it is
y(t) = e2t − 2e−3t.
(v)
ÿ − 4y = 0 with y(0) = 13 and ẏ(0) = 0.
We try y(t) = ekt then
k2 − 4k = 0
with solutions k = 0 or k = 4. The general solution is
y(t) = A + Be4t.
We have ẏ(t) = 4Be4t. The initial conditions
y(0) = A + B = 13 and ẏ(0) = 4B = 0
imply that A = 13 and B = 0, so the solution is
y(t) = 13.
(vi)
θ̈ + 4θ = 0 with θ(0) = 0 and θ̇(0) = 10.
With θ(t) = ekt we have
k2 + 4 = 0,
and so k = ±2i. The general solution is
θ(t) = A cos 2t + B sin 2t.
84 12 Homogeneous 2nd order linear ODEs
The derivative is